_-substitution: defining _ (more examples)

WEBVTT
Kind: captions
Language: en

00:00:00.613 --> 00:00:01.774
- [Instructor] What we're
going to do in this video

00:00:01.774 --> 00:00:03.933
is get some more practice
identifying when to use

00:00:03.933 --> 00:00:07.509
u-substitution and
picking an appropriate u.

00:00:07.509 --> 00:00:09.343
So let's say we have

00:00:09.343 --> 00:00:12.014
the indefinite integral of

00:00:12.014 --> 00:00:13.181
natural log of

00:00:14.661 --> 00:00:15.494
x to the,

00:00:17.842 --> 00:00:19.342
to the 10th power,

00:00:20.884 --> 00:00:22.967
all of that over x dx.

00:00:24.088 --> 00:00:25.481
Does u-substitution apply,

00:00:25.481 --> 00:00:28.639
and if so how would we
make that substitution?

00:00:28.639 --> 00:00:30.775
Well the key for u-substitution is to see,

00:00:30.775 --> 00:00:34.583
do I have some function
and its derivative?

00:00:34.583 --> 00:00:36.302
And you might immediately recognize

00:00:36.302 --> 00:00:38.020
that the derivative of natural log of x

00:00:38.020 --> 00:00:39.460
is equal to one over x.

00:00:39.460 --> 00:00:40.501
To make it a little bit clearer,

00:00:40.501 --> 00:00:43.845
I could write this as the integral of

00:00:43.845 --> 00:00:47.375
natural log of x to the 10th power

00:00:47.375 --> 00:00:49.292
times one over x dx.

00:00:50.788 --> 00:00:52.012
Now it becomes clear.

00:00:52.012 --> 00:00:54.404
We have some function, natural log of x,

00:00:54.404 --> 00:00:56.355
being raised to the 10th power,

00:00:56.355 --> 00:00:58.212
but we also have its
derivative right over here,

00:00:58.212 --> 00:00:59.512
one over x.

00:00:59.512 --> 00:01:01.718
So we could make the substitution.

00:01:01.718 --> 00:01:03.301
We could say that u

00:01:04.249 --> 00:01:06.455
is equal to the natural log of x.

00:01:06.455 --> 00:01:07.872
And the reason why I
picked natural log of x

00:01:07.872 --> 00:01:09.126
is because I see something,

00:01:09.126 --> 00:01:10.588
I see its exact derivative here,

00:01:10.588 --> 00:01:12.353
something close to its
derivative, in this case

00:01:12.353 --> 00:01:13.700
its exact derivative.

00:01:13.700 --> 00:01:15.720
And so then I could say,

00:01:15.720 --> 00:01:16.553
du dx,

00:01:17.508 --> 00:01:19.925
du dx is equal to one over x,

00:01:21.432 --> 00:01:23.015
which means that du

00:01:23.917 --> 00:01:26.334
is equal to one over x dx.

00:01:27.655 --> 00:01:29.071
And so here you have it.

00:01:29.071 --> 00:01:31.858
This right over here is du,

00:01:31.858 --> 00:01:35.712
and then this right over here is our u.

00:01:35.712 --> 00:01:38.290
And so this nicely simplifies to

00:01:38.290 --> 00:01:39.540
the integral of

00:01:40.496 --> 00:01:42.163
u to the 10th power,

00:01:44.606 --> 00:01:46.523
u to the 10th power du.

00:01:47.949 --> 00:01:49.946
And so you would evaluate what this is,

00:01:49.946 --> 00:01:51.525
find the antiderivative here,

00:01:51.525 --> 00:01:54.358
and then you would back-substitute
the natural log of x

00:01:54.358 --> 00:01:59.211
for u, to actually evaluate
this indefinite integral.

00:01:59.211 --> 00:02:00.720
Let's do another one.

00:02:00.720 --> 00:02:02.926
Let's say that we have

00:02:02.926 --> 00:02:04.259
the integral of,

00:02:06.223 --> 00:02:07.292
let's do something,

00:02:07.292 --> 00:02:10.287
let's do something interesting here.

00:02:10.287 --> 00:02:13.787
Let's say the integral of tangent x dx.

00:02:15.720 --> 00:02:18.739
Does u-substitution apply here?

00:02:18.739 --> 00:02:20.480
And at first you say well
I just have a tangent of x,

00:02:20.480 --> 00:02:22.570
where is its derivative?

00:02:22.570 --> 00:02:24.312
But one interesting thing to do is well

00:02:24.312 --> 00:02:27.354
we can rewrite tangent in
terms of sine and cosine.

00:02:27.354 --> 00:02:29.745
So we can write this as the integral of

00:02:29.745 --> 00:02:30.578
sine of x

00:02:31.789 --> 00:02:33.706
over cosine of x dx.

00:02:36.363 --> 00:02:37.361
And now you might say

00:02:37.361 --> 00:02:39.358
well where does u-substitution apply here?

00:02:39.358 --> 00:02:41.843
Well there's a couple of
ways to think about it.

00:02:41.843 --> 00:02:45.279
You could say the derivative
of sine of x is cosine of x,

00:02:45.279 --> 00:02:46.835
but you're now dividing by the derivative

00:02:46.835 --> 00:02:49.343
as opposed to multiplying by it.

00:02:49.343 --> 00:02:50.759
But more interesting you could say

00:02:50.759 --> 00:02:54.521
the derivative of cosine
of x is negative sine of x.

00:02:54.521 --> 00:02:56.166
We don't have a negative sine of x,

00:02:56.166 --> 00:02:57.978
but we can do a little bit of engineering.

00:02:57.978 --> 00:03:00.741
We can multiply by negative one twice.

00:03:00.741 --> 00:03:03.411
So we could say the negative
of the negative sine of x,

00:03:03.411 --> 00:03:05.640
and I stuck one of the you
could say negative ones

00:03:05.640 --> 00:03:06.801
outside of the integral,

00:03:06.801 --> 00:03:09.100
which comes straight from
our integration properties.

00:03:09.100 --> 00:03:10.331
This is equivalent.

00:03:10.331 --> 00:03:11.677
I can put a negative on the outside,

00:03:11.677 --> 00:03:12.908
a negative on the inside,

00:03:12.908 --> 00:03:15.950
so that this is the
derivative of cosine of x.

00:03:15.950 --> 00:03:17.831
And so now this is interesting.

00:03:17.831 --> 00:03:19.131
In fact let me rewrite this.

00:03:19.131 --> 00:03:21.058
This is going to be equal to negative,

00:03:21.058 --> 00:03:23.311
the negative integral,

00:03:23.311 --> 00:03:25.228
of one over cosine of x

00:03:27.676 --> 00:03:30.259
times negative sine of x dx.

00:03:33.690 --> 00:03:36.941
Now does it jump out at
you what our u might be?

00:03:36.941 --> 00:03:39.448
Well I have a cosine of
x in the denominator,

00:03:39.448 --> 00:03:41.770
and I have its derivative,

00:03:41.770 --> 00:03:45.462
so what if I made u equal to cosine of x?

00:03:45.462 --> 00:03:47.629
u is equal to cosine of x,

00:03:48.504 --> 00:03:49.671
and then du dx

00:03:51.848 --> 00:03:54.797
would be equal to negative sine of x.

00:03:54.797 --> 00:03:56.399
Or I could say

00:03:56.399 --> 00:04:00.149
that du is equal to negative sine of x dx.

00:04:01.740 --> 00:04:04.823
And just like that I have my du here,

00:04:05.687 --> 00:04:07.937
and this of course is my u.

00:04:09.843 --> 00:04:12.189
And so my whole thing
has now simplified to,

00:04:12.189 --> 00:04:14.162
it's equal to,

00:04:14.162 --> 00:04:15.162
the negative

00:04:16.461 --> 00:04:19.294
indefinite integral of one over u,

00:04:21.500 --> 00:04:22.667
one over u du.

00:04:25.215 --> 00:04:27.885
Which is a much easier
integral to evaluate,

00:04:27.885 --> 00:04:29.186
and then once you evaluate this,

00:04:29.186 --> 00:04:32.603
you back-substitute cosine of x for u.


Worked example: finding a Riemann sum using a table

WEBVTT
Kind: captions
Language: en

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- [Tutor] Imagine we're
asked to approximate the area

00:00:03.162 --> 00:00:06.421
between the x-axis and the graph of f

00:00:06.421 --> 00:00:09.488
from x equals one to x equals 10

00:00:09.488 --> 00:00:11.571
using a right Riemann sum

00:00:13.319 --> 00:00:16.367
with three equal subdivisions.

00:00:16.367 --> 00:00:20.686
To do that, we are given
a table of values for f,

00:00:20.686 --> 00:00:22.006
so I encourage you to pause the video

00:00:22.006 --> 00:00:24.612
and see if you can come
up with an approximation

00:00:24.612 --> 00:00:27.028
for the area between
the x axis and the graph

00:00:27.028 --> 00:00:29.367
from x equals one to x equals 10

00:00:29.367 --> 00:00:33.676
using a right Riemann sum
with three equal subdivisions,

00:00:33.676 --> 00:00:35.132
so I'm assuming you've had a go at it,

00:00:35.132 --> 00:00:37.073
so now let's try to do that together

00:00:37.073 --> 00:00:37.906
and this is interesting,

00:00:37.906 --> 00:00:40.128
because we don't have a
graph of the entire function,

00:00:40.128 --> 00:00:42.454
but we just have the value of the function

00:00:42.454 --> 00:00:43.445
at certain points,

00:00:43.445 --> 00:00:45.298
but as we'll see, this is all we need

00:00:45.298 --> 00:00:47.208
in order to get an
approximation for the area,

00:00:47.208 --> 00:00:48.584
we don't know how close it is

00:00:48.584 --> 00:00:50.490
to the actual area with just these points,

00:00:50.490 --> 00:00:52.208
but it'll give us at least

00:00:52.208 --> 00:00:54.833
a right Riemann sum for the approximation,

00:00:54.833 --> 00:00:57.338
or an approximation using
a right Riemann sum.

00:00:57.338 --> 00:00:59.381
So let me just draw some axes here,

00:00:59.381 --> 00:01:01.102
'cause whenever I do Riemann sums,

00:01:01.102 --> 00:01:02.963
you can do 'em without graphs,

00:01:02.963 --> 00:01:05.310
but it helps to think
about what's going on

00:01:05.310 --> 00:01:08.287
if you can visualize it graphically,

00:01:08.287 --> 00:01:12.686
so let's see, we are going from
x equals one to x equals 10,

00:01:12.686 --> 00:01:15.436
so this is one, two, three, four,

00:01:17.247 --> 00:01:19.997
five, six, seven, eight, nine, 10

00:01:22.673 --> 00:01:26.879
and so they give us the value
of f of x when x equals one,

00:01:26.879 --> 00:01:30.182
when x equals two, three, four,

00:01:30.182 --> 00:01:34.559
when x equals seven, five,
six, seven, eight, nine, 10

00:01:34.559 --> 00:01:38.706
and x equals 10 and they
tell us that when x is one,

00:01:38.706 --> 00:01:40.607
we're at six, then we go
to eight, to three, five,

00:01:40.607 --> 00:01:43.318
so let me mark these off, so
we're gonna go up to eight,

00:01:43.318 --> 00:01:45.901
so one, two, three, four, five,

00:01:48.464 --> 00:01:51.814
six, seven, eight and so what we know,

00:01:51.814 --> 00:01:55.193
when x is equal to one, f of one is six,

00:01:55.193 --> 00:01:58.443
so this is one, two, three, four, five,

00:01:59.554 --> 00:02:03.554
six, seven, eight, so
this point right over here

00:02:04.798 --> 00:02:08.342
is f of one, this is
the point one comma six

00:02:08.342 --> 00:02:11.685
and then we have the
point four comma eight,

00:02:11.685 --> 00:02:16.586
four comma eight, we'll
put this right about there

00:02:16.586 --> 00:02:19.939
and then we have seven
comma three is on our graph,

00:02:19.939 --> 00:02:22.992
y equals x f of x, so seven comma three

00:02:22.992 --> 00:02:25.681
would put us right over there

00:02:25.681 --> 00:02:29.299
and then we have 10 comma
five, so 10 comma five,

00:02:29.299 --> 00:02:31.052
put this right over there.

00:02:31.052 --> 00:02:32.732
That's all we know about the function,

00:02:32.732 --> 00:02:34.544
we don't know exactly what it looks like,

00:02:34.544 --> 00:02:37.377
our function might look like this,

00:02:38.877 --> 00:02:41.544
it might do something like this,

00:02:44.120 --> 00:02:46.438
whoops, I drew a part that
didn't look like a function,

00:02:46.438 --> 00:02:48.337
it might do something like this

00:02:48.337 --> 00:02:50.754
and oscillate really quickly,

00:02:51.802 --> 00:02:54.787
it might do, it might be nice and smooth

00:02:54.787 --> 00:02:58.340
and just kind of go and do
something just like that,

00:02:58.340 --> 00:03:01.457
kind of a connect the dots, we don't know,

00:03:01.457 --> 00:03:04.189
but we can still do the approximation

00:03:04.189 --> 00:03:07.011
using a right Riemann sum
with three equal subdivisions,

00:03:07.011 --> 00:03:08.078
how do we do that?

00:03:08.078 --> 00:03:09.634
Well, we're thinking about the area

00:03:09.634 --> 00:03:11.736
from x equals one to x equals 10,

00:03:11.736 --> 00:03:13.145
so let me make those boundaries clear,

00:03:13.145 --> 00:03:17.228
so this is from x equals
one to x equals 10

00:03:18.657 --> 00:03:22.382
and what we wanna do is have
three equal subdivisions

00:03:22.382 --> 00:03:25.292
and there's three very
natural subdivisions here,

00:03:25.292 --> 00:03:29.087
if we make each of our
subdivisions three-wide,

00:03:29.087 --> 00:03:32.062
so this could be a subdivision

00:03:32.062 --> 00:03:33.798
and then this is another subdivision

00:03:33.798 --> 00:03:34.847
and when you do Riemann sums,

00:03:34.847 --> 00:03:36.335
you don't have to have
three equal subdivisions,

00:03:36.335 --> 00:03:38.397
although that's what you'll often see,

00:03:38.397 --> 00:03:40.705
so we've just divided going from one to 10

00:03:40.705 --> 00:03:43.845
to three equal sections,
that are three-wide,

00:03:43.845 --> 00:03:46.950
so that's three, this is
three and this is three

00:03:46.950 --> 00:03:49.253
and so the question is how do we define

00:03:49.253 --> 00:03:51.347
the height of these subdivisions,

00:03:51.347 --> 00:03:52.887
which are going to end up being rectangles

00:03:52.887 --> 00:03:55.400
and that's where the
right Riemann sum applies,

00:03:55.400 --> 00:03:57.538
if we were doing a left Riemann sum,

00:03:57.538 --> 00:04:00.238
we would use the left boundary
of each of the subdivisions

00:04:00.238 --> 00:04:02.224
and the value of the function there

00:04:02.224 --> 00:04:03.632
to define the height of the rectangle,

00:04:03.632 --> 00:04:05.457
so this would be doing a left Riemann sum,

00:04:05.457 --> 00:04:07.653
but we're doing a right Riemann sum,

00:04:07.653 --> 00:04:09.997
so we use the right
boundary of each of these

00:04:09.997 --> 00:04:12.100
subdivisions to define the height,

00:04:12.100 --> 00:04:14.387
so our right boundary
is when x equals four

00:04:14.387 --> 00:04:17.527
for this first section, what
is f of four, it's eight,

00:04:17.527 --> 00:04:20.796
so we're gonna use that as the height

00:04:20.796 --> 00:04:23.345
of our, of this first rectangle,

00:04:23.345 --> 00:04:27.043
that's approximating the area
for this part of the curve.

00:04:27.043 --> 00:04:29.020
Similarly, for this second one,

00:04:29.020 --> 00:04:31.693
since we're using a right Riemann sum,

00:04:31.693 --> 00:04:34.472
we use the value of the
function at the right boundary,

00:04:34.472 --> 00:04:36.218
the right boundary is seven,

00:04:36.218 --> 00:04:38.256
so the value of the function is three,

00:04:38.256 --> 00:04:41.267
so this would be our second rectangle,

00:04:41.267 --> 00:04:45.197
our second division, I guess,
used to approximate the area

00:04:45.197 --> 00:04:46.224
and then last but not least,

00:04:46.224 --> 00:04:48.892
we would use the right boundary
of this third subdivision

00:04:48.892 --> 00:04:51.975
when x equals 10, f of 10 is five,

00:04:53.299 --> 00:04:54.740
so just like that

00:04:54.740 --> 00:04:57.787
and so then our right
Riemann approximation,

00:04:57.787 --> 00:05:00.694
using our right Riemann sum
with three equal subdivisions

00:05:00.694 --> 00:05:03.424
to approximate the area, we would just add

00:05:03.424 --> 00:05:06.636
the area of these rectangles,
so this first rectangle,

00:05:06.636 --> 00:05:10.295
let's see, it is three
wide and how high is it?

00:05:10.295 --> 00:05:14.581
Well, the height here is
f of four, which is eight,

00:05:14.581 --> 00:05:17.602
so this is going to be 24 square units,

00:05:17.602 --> 00:05:19.469
whatever the units happen to be,

00:05:19.469 --> 00:05:23.226
this is going to be three times
the height, here is three,

00:05:23.226 --> 00:05:26.833
f of seven is three, so
that is nine square units

00:05:26.833 --> 00:05:29.541
and then here, this is
three, the width is three

00:05:29.541 --> 00:05:32.498
times the height f of 10 is five,

00:05:32.498 --> 00:05:34.165
so three times five,

00:05:35.509 --> 00:05:37.664
which gets us 15

00:05:37.664 --> 00:05:39.439
and so our approximation of the area

00:05:39.439 --> 00:05:42.285
would be summing these three values up,

00:05:42.285 --> 00:05:44.990
so this would be 24 plus, let's see,

00:05:44.990 --> 00:05:49.578
nine plus 15 would give us
another 24, so it's 24 plus 24,

00:05:49.578 --> 00:05:50.995
it gets us to 48,

00:05:51.926 --> 00:05:53.882
there you go, we just
using that table of values,

00:05:53.882 --> 00:05:56.070
we've been able to find an approximation.

00:05:56.070 --> 00:05:58.625
Now once again, we don't know
how good our approximation is,

00:05:58.625 --> 00:06:00.824
it depends on what the function is doing,

00:06:00.824 --> 00:06:03.290
there is a world where it could
be very good approximation,

00:06:03.290 --> 00:06:06.724
maybe the function does
something like this,

00:06:06.724 --> 00:06:08.121
maybe the function just happens,

00:06:08.121 --> 00:06:09.667
let me make it a little bit,

00:06:09.667 --> 00:06:13.334
maybe the function does
something like this,

00:06:14.885 --> 00:06:16.930
where in this case, what we just did

00:06:16.930 --> 00:06:19.486
would be a very good approximation

00:06:19.486 --> 00:06:23.393
or maybe the function
does something like this,

00:06:23.393 --> 00:06:25.920
where in this situation,

00:06:25.920 --> 00:06:29.170
this might be a very bad approximation,

00:06:30.148 --> 00:06:32.793
but we can at least do the approximation

00:06:32.793 --> 00:06:36.793
using a right Riemann sum
just using this table.


Analyzing related rates problems: expressions

WEBVTT
Kind: captions
Language: en

00:00:00.734 --> 00:00:05.046
- [Instructor] The base b of
t of a triangle is decreasing

00:00:05.046 --> 00:00:07.713
at a rate of 13 meters per hour,

00:00:08.599 --> 00:00:12.864
and the height h of t of
the triangle is increasing

00:00:12.864 --> 00:00:16.104
at a rate of six meters per hour.

00:00:16.104 --> 00:00:18.984
At a certain instant t sub zero,

00:00:18.984 --> 00:00:23.509
the base is five meters and
the height is one meter.

00:00:23.509 --> 00:00:27.440
What is the rate of
change of the area A of t,

00:00:27.440 --> 00:00:29.704
so our area is going to be a function t,

00:00:29.704 --> 00:00:31.809
what is the rate of change of the area

00:00:31.809 --> 00:00:35.024
of the triangle at that instant?

00:00:35.024 --> 00:00:37.131
And so what we're going
to do in this exercise,

00:00:37.131 --> 00:00:39.402
instead of going straight
and trying to solve it,

00:00:39.402 --> 00:00:42.480
what we need to do here is
to identify the various units

00:00:42.480 --> 00:00:45.631
of different expressions
and then try to think about

00:00:45.631 --> 00:00:48.206
what information is given and what's not.

00:00:48.206 --> 00:00:50.577
And then that will actually equip us

00:00:50.577 --> 00:00:53.702
to actually solve this
rate of change problem.

00:00:53.702 --> 00:00:55.531
So let's just do this first part.

00:00:55.531 --> 00:00:58.266
Let's match each
expression with its units.

00:00:58.266 --> 00:00:59.883
And like always, pause the video,

00:00:59.883 --> 00:01:02.418
and see if you can do it on your own.

00:01:02.418 --> 00:01:06.412
All right, so the first
one is b prime of t.

00:01:06.412 --> 00:01:08.823
So this is the rate of change at which,

00:01:08.823 --> 00:01:12.100
of which the base is changing
with respect to time.

00:01:12.100 --> 00:01:15.908
So if we think about it,
b of t, that is the base.

00:01:15.908 --> 00:01:18.302
That is going to be in meters.

00:01:18.302 --> 00:01:21.015
So this is going to be in meters.

00:01:21.015 --> 00:01:22.932
If we say b prime of t,

00:01:24.213 --> 00:01:27.133
this is going to be how
much our base is changing

00:01:27.133 --> 00:01:28.566
with respect to time.

00:01:28.566 --> 00:01:32.147
So this is going to be meters per,

00:01:32.147 --> 00:01:33.506
and they give us right over here,

00:01:33.506 --> 00:01:36.691
they say it's decreasing at
a rate of 13 meters per hour,

00:01:36.691 --> 00:01:40.134
so the units here are meters per hour.

00:01:40.134 --> 00:01:44.301
And so b prime of t, that is
going to be in meters per hour.

00:01:45.432 --> 00:01:46.692
A at

00:01:46.692 --> 00:01:48.758
time t sub zero,

00:01:48.758 --> 00:01:51.770
remember, A is the area of our triangle.

00:01:51.770 --> 00:01:53.736
And we're measuring everything in meters.

00:01:53.736 --> 00:01:56.241
You can tell from the
information they've given us.

00:01:56.241 --> 00:01:59.502
And so area is going
to be in square units,

00:01:59.502 --> 00:02:02.919
and so it's going to be in square meters.

00:02:04.112 --> 00:02:07.435
Now, the height at time t sub zero,

00:02:07.435 --> 00:02:11.227
well, both the base and the
height, those are lengths.

00:02:11.227 --> 00:02:13.171
They're gonna be measured in meters,

00:02:13.171 --> 00:02:16.843
and so our height at
time t sub zero is going

00:02:16.843 --> 00:02:18.176
to be in meters.

00:02:19.122 --> 00:02:21.151
And then here we have the rate of change

00:02:21.151 --> 00:02:23.817
of our area with respect to time.

00:02:23.817 --> 00:02:27.050
So our area, we already
know is in meters squared.

00:02:27.050 --> 00:02:28.620
But we want know this here,

00:02:28.620 --> 00:02:30.536
that this is going to
be the rate of change

00:02:30.536 --> 00:02:32.480
of our area with respect to time.

00:02:32.480 --> 00:02:36.403
So it's going to be an
amount of area per unit time.

00:02:36.403 --> 00:02:39.049
And time here, we're using
hours, as you can see

00:02:39.049 --> 00:02:41.519
from some of the information
they've given us.

00:02:41.519 --> 00:02:44.545
So this is going to be area per unit time,

00:02:44.545 --> 00:02:47.135
or meters squared per hour.

00:02:47.135 --> 00:02:49.021
So it's going to be right over here.

00:02:49.021 --> 00:02:51.633
That's area per unit time.

00:02:51.633 --> 00:02:54.446
And the length we're
using in this is meters,

00:02:54.446 --> 00:02:56.117
and time is hours.

00:02:56.117 --> 00:02:57.371
All right.

00:02:57.371 --> 00:03:01.517
Now they say match each
expression with its given value.

00:03:01.517 --> 00:03:03.143
So what is

00:03:03.143 --> 00:03:05.793
the base of the triangle

00:03:05.793 --> 00:03:07.728
at time t sub zero?

00:03:07.728 --> 00:03:10.351
Do they give that to us?

00:03:10.351 --> 00:03:11.295
Well, let's see.

00:03:11.295 --> 00:03:13.019
They say, at a certain time,

00:03:13.019 --> 00:03:15.991
at a certain instant t sub zero, the base,

00:03:15.991 --> 00:03:17.767
I'm gonna underline this
in a different color,

00:03:17.767 --> 00:03:20.302
the base at a certain

00:03:20.302 --> 00:03:22.063
instant t sub zero,

00:03:22.063 --> 00:03:24.530
the base is five meters.

00:03:24.530 --> 00:03:27.373
So they say the base at time t sub zero,

00:03:27.373 --> 00:03:29.262
the base is a function of time,

00:03:29.262 --> 00:03:32.090
but they tell us that it is five meters.

00:03:32.090 --> 00:03:35.357
So this is five meters right over here.

00:03:35.357 --> 00:03:37.291
Now what about the rate of change

00:03:37.291 --> 00:03:39.724
of the base with respect to time?

00:03:39.724 --> 00:03:42.138
Do they tell us that?

00:03:42.138 --> 00:03:43.515
Well, look right over here.

00:03:43.515 --> 00:03:46.332
That's actually the first piece
of information they gave us.

00:03:46.332 --> 00:03:49.366
The base b of t of a
triangle is decreasing

00:03:49.366 --> 00:03:52.414
at a rate of 13 meters per hour.

00:03:52.414 --> 00:03:55.336
So the rate of change of the base,

00:03:55.336 --> 00:03:57.086
that it b prime of t,

00:03:58.439 --> 00:04:01.473
which is equal to db/dt,

00:04:01.473 --> 00:04:03.344
and they tell us that that is,

00:04:03.344 --> 00:04:06.629
it's decreasing at a rate
of 13 meters per hour.

00:04:06.629 --> 00:04:10.902
So that would be negative
13 meters per hour.

00:04:10.902 --> 00:04:14.187
And so the rate of change of
the base with respect to time

00:04:14.187 --> 00:04:16.734
is going to be negative 13.

00:04:16.734 --> 00:04:18.632
They gave us that.

00:04:18.632 --> 00:04:22.700
Now A prime of t, this is the
rate of change of the area

00:04:22.700 --> 00:04:24.283
at time t sub zero.

00:04:25.159 --> 00:04:27.294
Did they give us this?

00:04:27.294 --> 00:04:28.390
Well, they ask us that.

00:04:28.390 --> 00:04:32.419
What is the rate of
change of the area A of t

00:04:32.419 --> 00:04:34.276
of the triangle at that instant?

00:04:34.276 --> 00:04:36.628
So this is what we actually
need to figure out,

00:04:36.628 --> 00:04:37.942
but they haven't given it to us,

00:04:37.942 --> 00:04:39.916
otherwise there is no problem to solve.

00:04:39.916 --> 00:04:41.841
So this one right over here is not given.

00:04:41.841 --> 00:04:45.284
In fact, this is what we
are trying to solve for.

00:04:45.284 --> 00:04:48.039
And then finally we have

00:04:48.039 --> 00:04:51.398
the first derivative of the
height with respect to time.

00:04:51.398 --> 00:04:54.820
So you could view this as dh/dt.

00:04:54.820 --> 00:04:56.473
What is this going to be?

00:04:56.473 --> 00:04:58.362
Do they give it to us?

00:04:58.362 --> 00:04:59.641
Well, look right over here.

00:04:59.641 --> 00:05:03.689
They say the height of
the triangle is increasing

00:05:03.689 --> 00:05:06.439
at a rate of six meters per hour.

00:05:07.483 --> 00:05:10.348
So if they're saying h of t is increasing,

00:05:10.348 --> 00:05:11.859
they're telling us the rate of change

00:05:11.859 --> 00:05:14.115
of h of t with respect to time,

00:05:14.115 --> 00:05:15.821
so that's h prime of t.

00:05:15.821 --> 00:05:18.351
And they're telling us
that it is increasing

00:05:18.351 --> 00:05:19.583
at six meters per hour,

00:05:19.583 --> 00:05:22.610
so it's gonna be positive
six meters per hour.

00:05:22.610 --> 00:05:25.579
So they did indeed give us that.

00:05:25.579 --> 00:05:30.228
Now why is all of this a
useful exercise to go through?

00:05:30.228 --> 00:05:34.480
Well, now we are really
ready to solve the question.

00:05:34.480 --> 00:05:36.955
Because, in general, if we're
talking about any triangle,

00:05:36.955 --> 00:05:39.041
we know that area

00:05:39.041 --> 00:05:43.238
is equal to 1/2 base times height.

00:05:43.238 --> 00:05:47.236
Now in this situation, area
and our base and our height,

00:05:47.236 --> 00:05:48.869
they're all going to be functions of t,

00:05:48.869 --> 00:05:50.869
so we could write A of t

00:05:51.846 --> 00:05:55.096
is equal to 1/2 times b of t

00:05:56.012 --> 00:05:57.095
times h of t.

00:05:58.632 --> 00:06:02.033
And if we want to find the rate of change

00:06:02.033 --> 00:06:04.106
of our area at that instant,

00:06:04.106 --> 00:06:05.792
and the instant that they're talking about

00:06:05.792 --> 00:06:07.766
is at time t sub zero,

00:06:07.766 --> 00:06:10.026
well, then what we would want
to do is take the derivative

00:06:10.026 --> 00:06:12.099
of both sides with respect to t.

00:06:12.099 --> 00:06:13.677
So the derivative on the left-hand side

00:06:13.677 --> 00:06:17.275
with respect to t would be A prime of t,

00:06:17.275 --> 00:06:20.667
and then on the right-hand
side it would be 1/2 times,

00:06:20.667 --> 00:06:23.652
and we would actually
use a combination of,

00:06:23.652 --> 00:06:26.022
well, it's really just the
product rule right over here.

00:06:26.022 --> 00:06:28.679
The derivative of the first
function with respect to t,

00:06:28.679 --> 00:06:32.292
so it's b prime of t
times the second function,

00:06:32.292 --> 00:06:34.251
this is just the product rule here,

00:06:34.251 --> 00:06:36.751
plus the first function b of t

00:06:38.067 --> 00:06:40.041
times the derivative
of the second function

00:06:40.041 --> 00:06:41.977
with respect to time.

00:06:41.977 --> 00:06:45.438
And we need to figure out not
just the general expression,

00:06:45.438 --> 00:06:48.799
they want us to know what the
rate of change of the area,

00:06:48.799 --> 00:06:52.578
so A prime of t at that
instant, at t sub zero.

00:06:52.578 --> 00:06:54.538
So what we want to figure out,

00:06:54.538 --> 00:06:58.318
we want to figure out A
prime at time t sub zero.

00:06:58.318 --> 00:07:02.032
Well, that's just going to
be equal to 1/2 times b prime

00:07:02.032 --> 00:07:05.032
of t sub zero times h of t sub zero,

00:07:06.001 --> 00:07:08.578
plus b of t sub zero,

00:07:08.578 --> 00:07:11.147
times h prime of t sub zero.

00:07:11.147 --> 00:07:12.601
Now this might seem daunting,

00:07:12.601 --> 00:07:15.627
except they've given us a
lot of this information.

00:07:15.627 --> 00:07:17.578
What is b prime

00:07:17.578 --> 00:07:18.745
of t sub zero?

00:07:19.859 --> 00:07:21.523
Well, they tell us the rate of change

00:07:21.523 --> 00:07:22.872
of b with respect to time,

00:07:22.872 --> 00:07:24.864
and it seems like it's just gonna stay

00:07:24.864 --> 00:07:27.665
at negative 13 meters per hour.

00:07:27.665 --> 00:07:29.537
So they gave us this.

00:07:29.537 --> 00:07:32.698
H, what is the height at time t sub zero?

00:07:32.698 --> 00:07:35.615
Well, they tell us right over here.

00:07:36.598 --> 00:07:39.308
At a certain instant,
the base is five meters

00:07:39.308 --> 00:07:40.488
and the height is one meter.

00:07:40.488 --> 00:07:43.450
So they give us both
b and h at t sub zero.

00:07:43.450 --> 00:07:44.885
So they gave us this.

00:07:44.885 --> 00:07:46.096
They gave us this.

00:07:46.096 --> 00:07:47.801
And what is the rate
of change of the height

00:07:47.801 --> 00:07:49.316
at time t sub zero?

00:07:49.316 --> 00:07:50.505
Well, they tell us.

00:07:50.505 --> 00:07:52.722
The height of the triangle is increasing

00:07:52.722 --> 00:07:54.725
at a rate of six meters per hour.

00:07:54.725 --> 00:07:57.476
So they tell us that as well.

00:07:57.476 --> 00:07:58.931
All of that stuff is given,

00:07:58.931 --> 00:08:00.906
and so you just have to
plug it in to figure out

00:08:00.906 --> 00:08:04.456
what is the rate of change
of the area at t sub zero,

00:08:04.456 --> 00:08:05.789
at that instant.


Mistakes when finding inflection points: not checking candidates

WEBVTT
Kind: captions
Language: en

00:00:00.477 --> 00:00:02.215
- [Instructor] Olga was asked to find

00:00:02.215 --> 00:00:05.263
where f of x is equal to x
minus two to the fourth power

00:00:05.263 --> 00:00:06.667
has inflection points.

00:00:06.667 --> 00:00:08.379
This is her solution.

00:00:08.379 --> 00:00:10.134
So we look at her solution
and then they ask us,

00:00:10.134 --> 00:00:12.170
is Olga's work correct?

00:00:12.170 --> 00:00:14.292
If not, what's her mistake?

00:00:14.292 --> 00:00:17.428
So pause this video and see
if you can figure this out.

00:00:17.428 --> 00:00:19.219
Alright, let's just follow her work.

00:00:19.219 --> 00:00:21.664
So here, she's trying to
take the first derivative.

00:00:21.664 --> 00:00:23.093
So you would apply the chain rule.

00:00:23.093 --> 00:00:26.230
It would be four times x
minus two to the third power

00:00:26.230 --> 00:00:29.165
times the derivative of x
minus two which is just one.

00:00:29.165 --> 00:00:30.998
So this checks out.

00:00:30.998 --> 00:00:32.265
Then you take the derivative of this.

00:00:32.265 --> 00:00:34.340
It'd be three times four which would be 12

00:00:34.340 --> 00:00:36.496
times x minus two to the second power

00:00:36.496 --> 00:00:38.694
times the derivative of x
minus two which is just one

00:00:38.694 --> 00:00:40.360
which is exactly what she has here,

00:00:40.360 --> 00:00:42.848
12 times x minus two to the second power.

00:00:42.848 --> 00:00:46.151
That checks out so step
one's looking good for Olga.

00:00:46.151 --> 00:00:49.539
Step two, the solution
of the second derivative

00:00:49.539 --> 00:00:51.946
equaling zero is x equals two.

00:00:51.946 --> 00:00:53.046
That looks right.

00:00:53.046 --> 00:00:56.835
The second derivative is 12
times x minus two squared

00:00:56.835 --> 00:00:58.992
and we wanna make that equal to zero.

00:00:58.992 --> 00:01:02.623
This is only going to be
true when x is equal to two.

00:01:02.623 --> 00:01:05.155
So step two is looking good.

00:01:05.155 --> 00:01:09.171
So step three, Olga says
f has an inflection point

00:01:09.171 --> 00:01:10.743
at x equals two.

00:01:10.743 --> 00:01:13.343
So she's basing this just on the fact

00:01:13.343 --> 00:01:16.364
that the second derivative is
zero when x is equal to two.

00:01:16.364 --> 00:01:18.205
She's basing this just on the fact

00:01:18.205 --> 00:01:21.513
that f prime prime of
two is equal to zero.

00:01:21.513 --> 00:01:23.391
Now, I have a problem with this

00:01:23.391 --> 00:01:25.593
because the fact that
your second derivative

00:01:25.593 --> 00:01:26.983
is zero at x equals two,

00:01:26.983 --> 00:01:29.594
that makes two a nice
candidate to check out

00:01:29.594 --> 00:01:31.224
but you can't immediately say

00:01:31.224 --> 00:01:33.591
that we have an inflection point there.

00:01:33.591 --> 00:01:35.657
Remember, an inflection
point is where we go

00:01:35.657 --> 00:01:38.194
from being concave upwards
to concave downwards

00:01:38.194 --> 00:01:40.475
or concave downwards to concave upwards

00:01:40.475 --> 00:01:42.634
and speaking in the language
of the second derivative,

00:01:42.634 --> 00:01:45.689
it means that the second
derivative changes signs

00:01:45.689 --> 00:01:50.336
as we go from below x equals
two to above x equals two

00:01:50.336 --> 00:01:51.560
but we have to test that

00:01:51.560 --> 00:01:54.123
because it's not
necessarily always the case.

00:01:54.123 --> 00:01:56.159
So let's actually test it.

00:01:56.159 --> 00:01:59.056
Let's think about some intervals.

00:01:59.056 --> 00:01:59.889
Intervals.

00:02:00.888 --> 00:02:02.519
So let's think about the interval

00:02:02.519 --> 00:02:05.130
when we go from negative infinity to two

00:02:05.130 --> 00:02:06.069
and let's think about the interval

00:02:06.069 --> 00:02:08.636
where we go from two to positive infinity.

00:02:08.636 --> 00:02:13.163
If you want, you could
have some test values.

00:02:13.163 --> 00:02:16.137
You could think about the sign,

00:02:16.137 --> 00:02:18.554
sign of our second derivative

00:02:20.063 --> 00:02:23.976
and then based on that, you
could think about concavity,

00:02:23.976 --> 00:02:25.226
concavity of f.

00:02:28.093 --> 00:02:29.966
So let's think about what's happening.

00:02:29.966 --> 00:02:31.312
So you could take a test value.

00:02:31.312 --> 00:02:33.474
Let's say one is in this interval

00:02:33.474 --> 00:02:35.309
and let's say three is in this interval

00:02:35.309 --> 00:02:39.637
and you could say one minus
two squared is going to be,

00:02:39.637 --> 00:02:41.819
let's see, that's negative
one squared which is one

00:02:41.819 --> 00:02:44.585
and then you're just going to,
this is just going to be 12

00:02:44.585 --> 00:02:48.090
so this is going to be positive
and if you tried three,

00:02:48.090 --> 00:02:50.781
three minus two squared is one times 12.

00:02:50.781 --> 00:02:52.534
Well, that's also going to be positive

00:02:52.534 --> 00:02:55.591
and so you're going to be concave upwards.

00:02:55.591 --> 00:02:57.177
At least in these test values,

00:02:57.177 --> 00:03:00.002
it looks like on either side of two

00:03:00.002 --> 00:03:01.291
that the sign of the second derivative

00:03:01.291 --> 00:03:03.163
is positive on either side
of the two and you might say,

00:03:03.163 --> 00:03:05.892
well, maybe I just need
to find closer values

00:03:05.892 --> 00:03:08.094
but if you inspect the
second derivative here,

00:03:08.094 --> 00:03:10.335
you can see that this is
never going to be negative.

00:03:10.335 --> 00:03:14.028
In fact, for any value
other than x equals two

00:03:14.028 --> 00:03:15.930
and this value right
over here since we're,

00:03:15.930 --> 00:03:18.868
even if x minus two is
negative, you're squaring it

00:03:18.868 --> 00:03:20.660
which will make this entire thing positive

00:03:20.660 --> 00:03:22.899
and then multiplying it
times a positive value.

00:03:22.899 --> 00:03:26.409
So for any value other than x equals two,

00:03:26.409 --> 00:03:28.860
the sign of our second
derivative is positive

00:03:28.860 --> 00:03:32.319
which means that we're
going to be concave upwards

00:03:32.319 --> 00:03:34.558
and so we actually don't
have an inflection point

00:03:34.558 --> 00:03:37.207
at x equals two because
we are not switching signs

00:03:37.207 --> 00:03:41.119
as we go from values
less than x equals two

00:03:41.119 --> 00:03:43.518
to values greater than x equals two.

00:03:43.518 --> 00:03:46.820
Our second derivative
is not switching sides.

00:03:46.820 --> 00:03:48.781
So once again, this is incorrect.

00:03:48.781 --> 00:03:51.349
We actually don't have an
inflection point at x equals two

00:03:51.349 --> 00:03:54.077
because our second derivative
does not switch signs

00:03:54.077 --> 00:03:55.831
as we cross x equals two

00:03:55.831 --> 00:03:59.331
which means our concavity does not change.


Mistakes when finding inflection points: second derivative undefined

WEBVTT
Kind: captions
Language: en

00:00:00.640 --> 00:00:03.149
- [Instructor] Robert was
asked to find where g of x,

00:00:03.149 --> 00:00:06.198
which is equal to the cube root
of x, has inflection points.

00:00:06.198 --> 00:00:08.295
This is his solution.

00:00:08.295 --> 00:00:12.267
And then later we are asked,
"Is Robert's work correct?

00:00:12.267 --> 00:00:14.460
"If not, what's his mistake?"

00:00:14.460 --> 00:00:18.332
So pause this video and try
to figure it out on your own.

00:00:18.332 --> 00:00:19.903
All right, now let's work
through this together.

00:00:19.903 --> 00:00:24.204
So our original g of x is
equal to the cube root of x,

00:00:24.204 --> 00:00:26.991
which is the same thing as x to the 1/3.

00:00:26.991 --> 00:00:28.848
So in step one it looks
like Robert's trying

00:00:28.848 --> 00:00:30.777
to find the first and second derivative.

00:00:30.777 --> 00:00:32.554
So the first derivative,
we just do the power rule,

00:00:32.554 --> 00:00:35.089
so it'll be 1/3x to the
decrement of the exponent,

00:00:35.089 --> 00:00:36.990
so this is looking good.

00:00:36.990 --> 00:00:40.950
Second derivative, we take
this, multiply this times 1/3,

00:00:40.950 --> 00:00:42.651
which would be negative 2/9.

00:00:42.651 --> 00:00:44.385
And then decrement, negative 2/3,

00:00:44.385 --> 00:00:47.784
which would indeed by negative
5/3, so that looks right.

00:00:47.784 --> 00:00:49.830
And then it looks like
Robert's trying to rewrite it.

00:00:49.830 --> 00:00:52.114
So we have the negative 2/9 still,

00:00:52.114 --> 00:00:54.955
but then he recognized
that this is the same thing

00:00:54.955 --> 00:00:57.531
as x to 5/3 in the denominator,

00:00:57.531 --> 00:00:59.931
and x to the 5/3 is the same thing

00:00:59.931 --> 00:01:02.954
as the cube root of x to the fifth.

00:01:02.954 --> 00:01:04.107
So this is all looking good.

00:01:04.107 --> 00:01:06.464
Step one looks good.

00:01:06.464 --> 00:01:08.503
And then step two, it
looks like he's trying

00:01:08.503 --> 00:01:10.962
to find the solution or
he's trying to find x values

00:01:10.962 --> 00:01:13.377
where the second derivative
is equal to zero,

00:01:13.377 --> 00:01:17.215
and it is indeed true
that this has no solution,

00:01:17.215 --> 00:01:19.680
that you can never make
this second derivative

00:01:19.680 --> 00:01:20.558
equal to zero.

00:01:20.558 --> 00:01:22.621
In order to be zero, the
numerator would have to be zero.

00:01:22.621 --> 00:01:25.347
And, well, two is never
going to be equal to zero.

00:01:25.347 --> 00:01:27.724
So this is correct.

00:01:27.724 --> 00:01:29.114
And then step three,

00:01:29.114 --> 00:01:32.763
he says g doesn't have
any inflection points.

00:01:32.763 --> 00:01:35.354
Now, this is a little bit suspect.

00:01:35.354 --> 00:01:38.797
It is in many cases our
inflection point is a situation

00:01:38.797 --> 00:01:41.861
where our second derivative
is equal to zero,

00:01:41.861 --> 00:01:44.007
and even then we don't know
it's an inflection point.

00:01:44.007 --> 00:01:45.560
It would be a candidate inflection point.

00:01:45.560 --> 00:01:46.773
We would have to confirm

00:01:46.773 --> 00:01:49.063
that our second derivative crosses signs

00:01:49.063 --> 00:01:51.980
or switches signs as
we cross that x value.

00:01:51.980 --> 00:01:53.790
But here we can't find a situation

00:01:53.790 --> 00:01:55.767
where our second derivative
is equal to zero,

00:01:55.767 --> 00:01:57.543
but we have to remind ourselves

00:01:57.543 --> 00:01:59.489
that other candidate inflection points

00:01:59.489 --> 00:02:02.905
are where our second
derivative is undefined.

00:02:02.905 --> 00:02:05.868
And so he can't make this
statement without seeing

00:02:05.868 --> 00:02:08.816
where our second derivative
could be undefined.

00:02:08.816 --> 00:02:11.347
So, for example, he could say

00:02:11.347 --> 00:02:14.847
that g prime prime is undefined when what?

00:02:18.751 --> 00:02:20.249
Well, this is going to be undefined

00:02:20.249 --> 00:02:22.091
when x is equal to zero.

00:02:22.091 --> 00:02:24.355
Zero to the fifth, cube root
of that, that's gonna be zero.

00:02:24.355 --> 00:02:26.256
But then you're dividing by zero.

00:02:26.256 --> 00:02:29.746
So g prime prime undefined
when x is equal to zero.

00:02:29.746 --> 00:02:33.913
So therefore x equals,
so we could say candidate

00:02:38.498 --> 00:02:41.748
inflection point when x equals zero.

00:02:46.590 --> 00:02:49.157
And so then we would want to test it.

00:02:49.157 --> 00:02:51.133
And we could set up a traditional table

00:02:51.133 --> 00:02:52.835
that you might have seen before

00:02:52.835 --> 00:02:56.168
where we have our interval or intervals.

00:02:58.022 --> 00:03:01.758
We could have test
values in our intervals.

00:03:01.758 --> 00:03:03.091
We have to be careful with those.

00:03:03.091 --> 00:03:04.829
Make sure that they are indicative.

00:03:04.829 --> 00:03:07.329
And then we would say the sign

00:03:08.348 --> 00:03:11.992
of our second derivative of g prime prime.

00:03:11.992 --> 00:03:15.084
And then we would have our concavity.

00:03:15.084 --> 00:03:16.334
Concavity of g.

00:03:17.992 --> 00:03:20.827
And in order for x equals zero
to be an inflection point,

00:03:20.827 --> 00:03:22.877
we would have to switch signs,

00:03:22.877 --> 00:03:25.216
or our second derivative
would have to switch signs

00:03:25.216 --> 00:03:27.285
as we cross x equals zero,

00:03:27.285 --> 00:03:29.905
which would mean our
concavity of g switches signs

00:03:29.905 --> 00:03:32.393
as we cross x equals zero.

00:03:32.393 --> 00:03:34.159
So let's do values less than zero,

00:03:34.159 --> 00:03:35.910
negative infinity to zero

00:03:35.910 --> 00:03:39.954
and then values greater
than zero, zero to infinity.

00:03:39.954 --> 00:03:41.231
I could do test values.

00:03:41.231 --> 00:03:44.586
Let's say, I'll use negative one and one.

00:03:44.586 --> 00:03:46.192
And you have to be careful
when you use these.

00:03:46.192 --> 00:03:48.565
You have to make sure
that we are close enough

00:03:48.565 --> 00:03:53.127
that nothing unusual happens
between these test values

00:03:53.127 --> 00:03:56.515
up until we get to that
candidate inflection point.

00:03:56.515 --> 00:03:58.263
And, now, what's the sign
of our second derivative

00:03:58.263 --> 00:04:00.215
when x is equal to negative one?

00:04:00.215 --> 00:04:02.032
When x equals negative one.

00:04:02.032 --> 00:04:03.556
So, let's see, negative one

00:04:03.556 --> 00:04:06.150
to the fifth power is negative one.

00:04:06.150 --> 00:04:09.149
Cube root of negative one is negative one.

00:04:09.149 --> 00:04:11.590
And so we're gonna have negative 2/9

00:04:11.590 --> 00:04:12.477
divided by negative one.

00:04:12.477 --> 00:04:14.269
It's gonna be positive 2/9.

00:04:14.269 --> 00:04:18.237
So our sign right over
here is gonna be positive.

00:04:18.237 --> 00:04:19.556
And this is gonna be in general

00:04:19.556 --> 00:04:21.305
when we're dealing with
any negative value,

00:04:21.305 --> 00:04:23.280
'cause if you take any negative
value to the fifth power,

00:04:23.280 --> 00:04:24.535
it's gonna be negative.

00:04:24.535 --> 00:04:26.055
And you take the cube root of that

00:04:26.055 --> 00:04:27.263
and you're gonna have negative.

00:04:27.263 --> 00:04:28.967
But then you have a negative
value divided by that.

00:04:28.967 --> 00:04:30.233
You're gonna get a positive value.

00:04:30.233 --> 00:04:32.105
So you can feel good that
this test value's indicative

00:04:32.105 --> 00:04:33.935
of actually this entire interval.

00:04:33.935 --> 00:04:35.934
And if you're dealing
with a positive value,

00:04:35.934 --> 00:04:37.908
well, that to the fifth
power is gonna be positive.

00:04:37.908 --> 00:04:40.420
Cube root of that is still
going to be positive.

00:04:40.420 --> 00:04:41.746
But then you're gonna have negative 2/9

00:04:41.746 --> 00:04:43.357
divided by that positive value,

00:04:43.357 --> 00:04:45.192
so this is going to be negative.

00:04:45.192 --> 00:04:48.843
So it is indeed the case that
our concavity of g switches

00:04:48.843 --> 00:04:50.285
as we cross x equals zero.

00:04:50.285 --> 00:04:53.879
We're concave upwards
when x is less than zero.

00:04:53.879 --> 00:04:55.884
Our second derivative is positive.

00:04:55.884 --> 00:05:00.693
And we're concave downwards
when x is greater than zero.

00:05:00.693 --> 00:05:02.084
Let me write that a little bit.

00:05:02.084 --> 00:05:03.670
Downwards.

00:05:03.670 --> 00:05:07.087
Downwards when x is greater than zero.

00:05:08.778 --> 00:05:12.815
So we are switching concavity
as we cross x equals zero,

00:05:12.815 --> 00:05:15.828
and so this tells us that x.

00:05:15.828 --> 00:05:18.766
So let's see, we are switching signs.

00:05:18.766 --> 00:05:20.210
Switching.

00:05:20.210 --> 00:05:23.543
Let me say g prime prime switching signs

00:05:24.536 --> 00:05:26.953
as we cross x equals zero.

00:05:30.237 --> 00:05:32.967
And our function is
defined at x equals zero.

00:05:32.967 --> 00:05:36.384
And function defined at x equals zero.

00:05:40.773 --> 00:05:45.251
So we have an inflection
point at x equals zero.

00:05:45.251 --> 00:05:49.001
So inflection point at x is equal to zero.

00:05:53.867 --> 00:05:57.276
And if you're familiar with
the graph of the cube root,

00:05:57.276 --> 00:06:02.006
you would indeed see an
inflection point at that point.

00:06:02.006 --> 00:06:03.204
So there we go.

00:06:03.204 --> 00:06:04.299
He was wrong in step three.

00:06:04.299 --> 00:06:05.634
There actually is an inflection point.

00:06:05.634 --> 00:06:07.545
It's not when the second
derivative is equal to zero.

00:06:07.545 --> 00:06:10.703
It's actually where the second
derivative is undefined.


Justification using second derivative: maximum point

WEBVTT
Kind: captions
Language: en

00:00:00.516 --> 00:00:02.439
- [Instructor] We're told
that given that h prime

00:00:02.439 --> 00:00:04.933
of negative four is equal to zero,

00:00:04.933 --> 00:00:08.864
what is an appropriate
calculus-based justification

00:00:08.864 --> 00:00:12.490
for the fact that h has a relative maximum

00:00:12.490 --> 00:00:15.461
at x is equal to negative four?

00:00:15.461 --> 00:00:16.811
So right over here,

00:00:16.811 --> 00:00:19.934
we actually have the
graph of our function h.

00:00:19.934 --> 00:00:22.144
This is the graph y is equal to h of x.

00:00:22.144 --> 00:00:24.151
And we don't have graphed
the first derivative,

00:00:24.151 --> 00:00:26.178
but we do have graphed
the second derivative

00:00:26.178 --> 00:00:29.578
right here in this orange
color, h prime prime.

00:00:29.578 --> 00:00:30.565
So they're telling us,

00:00:30.565 --> 00:00:33.493
given that h prime of negative
four is equal to zero,

00:00:33.493 --> 00:00:35.907
so that's saying that given
that the first derivative

00:00:35.907 --> 00:00:37.811
at x equals negative
four is equal to zero,

00:00:37.811 --> 00:00:39.756
and you can see that the
slope of the tangent line

00:00:39.756 --> 00:00:44.750
when x is equal to negative
four does indeed equal zero.

00:00:44.750 --> 00:00:48.435
So given that, what is a calculus-based,

00:00:48.435 --> 00:00:52.092
let me underline that, a
calculus-based justification

00:00:52.092 --> 00:00:54.454
for the fact that h has a relative maximum

00:00:54.454 --> 00:00:56.611
at x equals negative four?

00:00:56.611 --> 00:00:59.239
So this first one says
that the second derivative

00:00:59.239 --> 00:01:02.424
at x equals negative four is negative.

00:01:02.424 --> 00:01:04.108
Now, what does that tell us?

00:01:04.108 --> 00:01:07.315
If the second derivative is negative,

00:01:07.315 --> 00:01:12.276
that means that the first
derivative is decreasing,

00:01:12.276 --> 00:01:14.730
which is another way of
saying that we are dealing

00:01:14.730 --> 00:01:18.219
with a situation where, at
least at x equals negative four,

00:01:18.219 --> 00:01:20.302
we are concave downwards,

00:01:21.903 --> 00:01:25.598
downwards, which means that
the general shape of our curve

00:01:25.598 --> 00:01:27.396
is going to look something like this

00:01:27.396 --> 00:01:29.507
around x equals negative four.

00:01:29.507 --> 00:01:33.423
And if the slope at x equals
negative four is zero,

00:01:33.423 --> 00:01:36.340
well, that tells us that,
yes, we indeed are dealing

00:01:36.340 --> 00:01:38.809
with a relative maximum point.

00:01:38.809 --> 00:01:41.856
If the second derivative
at that point was positive,

00:01:41.856 --> 00:01:43.917
then we would be concave upwards.

00:01:43.917 --> 00:01:46.215
And then if our derivative is zero there,

00:01:46.215 --> 00:01:49.202
we'd say, okay, that's a
relative minimum point.

00:01:49.202 --> 00:01:51.671
But this is indeed true.

00:01:51.671 --> 00:01:55.091
The second derivative is negative
at x equals negative four,

00:01:55.091 --> 00:01:57.035
which means we are concave downwards,

00:01:57.035 --> 00:01:59.952
which means that we are a upside U,

00:02:00.972 --> 00:02:03.194
and that point where
the derivative is zero

00:02:03.194 --> 00:02:04.749
is indeed a relative maximum.

00:02:04.749 --> 00:02:06.526
So let me, so that is the answer.

00:02:06.526 --> 00:02:09.519
And we're done, but let's
just rule out the other ones.

00:02:09.519 --> 00:02:12.850
H increase before x equals negative four.

00:02:12.850 --> 00:02:14.075
That is indeed true.

00:02:14.075 --> 00:02:16.705
Before x equals negative
four, we are increasing,

00:02:16.705 --> 00:02:18.281
and h decreases after it.

00:02:18.281 --> 00:02:21.890
That is true, and that is one
rationale for thinking that,

00:02:21.890 --> 00:02:24.296
hey, we must have a maximum point,

00:02:24.296 --> 00:02:26.208
assuming that our function is continuous

00:02:26.208 --> 00:02:28.290
at x equals negative four.

00:02:28.290 --> 00:02:29.123
So this is true.

00:02:29.123 --> 00:02:31.439
It is a justification
for a relative maximum,

00:02:31.439 --> 00:02:33.718
but it is not calculus-based.

00:02:33.718 --> 00:02:37.244
And so that's why we
can rule this one out.

00:02:37.244 --> 00:02:39.590
The second derivative
has a relative minimum

00:02:39.590 --> 00:02:41.465
at x equals negative four.

00:02:41.465 --> 00:02:43.729
Well, it does indeed seem to be true.

00:02:43.729 --> 00:02:45.366
There's a relative minimum there,

00:02:45.366 --> 00:02:48.461
but that's not a justification for why

00:02:48.461 --> 00:02:51.761
this is why h of negative four

00:02:51.761 --> 00:02:55.448
or why we have a relative maximum
at x equals negative four.

00:02:55.448 --> 00:02:57.405
For example, this,

00:02:57.405 --> 00:03:01.631
you could have a relative minimum
in your second derivative,

00:03:01.631 --> 00:03:04.030
but your second derivative
could still be positive there.

00:03:04.030 --> 00:03:05.761
So what if the second
derivative was like that?

00:03:05.761 --> 00:03:07.681
That would still be a relative minimum.

00:03:07.681 --> 00:03:09.671
But if it was positive at that point,

00:03:09.671 --> 00:03:11.795
then you would be concave upwards,

00:03:11.795 --> 00:03:14.050
which would mean that at
x equals negative four,

00:03:14.050 --> 00:03:16.345
your original function
wouldn't have a maximum point,

00:03:16.345 --> 00:03:18.097
it would have a minimum point.

00:03:18.097 --> 00:03:20.808
And so just a relative
minimum isn't enough.

00:03:20.808 --> 00:03:22.296
In order to know that you are dealing

00:03:22.296 --> 00:03:24.834
with a relative maximum,
you would have to know

00:03:24.834 --> 00:03:27.529
that the second derivative
is negative there.

00:03:27.529 --> 00:03:31.234
Now, this fourth choice, h
prime prime is concave up,

00:03:31.234 --> 00:03:33.907
it does indeed look like
the second derivative

00:03:33.907 --> 00:03:38.515
is concave up, but that,
by itself, does not justify

00:03:38.515 --> 00:03:41.735
that the original function is concave up.

00:03:41.735 --> 00:03:44.713
For example, well, I could
use this example right here.

00:03:44.713 --> 00:03:47.341
This is a potential second derivative

00:03:47.341 --> 00:03:49.322
that is concave upwards,

00:03:49.322 --> 00:03:51.527
but it is positive the entire time.

00:03:51.527 --> 00:03:54.519
And if your second derivative
is positive the entire time,

00:03:54.519 --> 00:03:56.174
that means that your first derivative

00:03:56.174 --> 00:03:58.154
is increasing the entire time,

00:03:58.154 --> 00:04:00.670
which means that your original function

00:04:00.670 --> 00:04:04.159
is going to be concave
upwards the entire time.

00:04:04.159 --> 00:04:06.725
And so if you're concave
upwards the entire time,

00:04:06.725 --> 00:04:08.859
then you would not have a relative maximum

00:04:08.859 --> 00:04:10.443
at x equals negative four.

00:04:10.443 --> 00:04:13.140
So we would rule that one out as well.


Calculus based justification for function increasing

WEBVTT
Kind: captions
Language: en

00:00:00.388 --> 00:00:02.885
- [Instructor] We are told
the differentiable function h

00:00:02.885 --> 00:00:05.210
and its derivative h prime are graphed

00:00:05.210 --> 00:00:07.214
and you can see it here, h is in blue

00:00:07.214 --> 00:00:10.560
and then its derivative h
prime is in this orange color.

00:00:10.560 --> 00:00:12.278
Four students were asked to give

00:00:12.278 --> 00:00:15.370
an appropriate
calculus-based justification

00:00:15.370 --> 00:00:17.288
for the fact that h is increasing

00:00:17.288 --> 00:00:19.491
when x is greater than zero.

00:00:19.491 --> 00:00:23.074
Can you match the teacher's
comments to the justifications?

00:00:23.074 --> 00:00:25.846
So before I even look at
what the students wrote,

00:00:25.846 --> 00:00:28.124
you might say, hey, look,
I can just look at this

00:00:28.124 --> 00:00:31.701
and see that h is increasing
when x is greater than zero

00:00:31.701 --> 00:00:33.613
but just by looking at the graph of h,

00:00:33.613 --> 00:00:36.139
that by itself is not a
calculus-based justification.

00:00:36.139 --> 00:00:37.281
We're not using calculus.

00:00:37.281 --> 00:00:38.342
We're just using our knowledge

00:00:38.342 --> 00:00:40.461
of what it means for a
graph to be increasing.

00:00:40.461 --> 00:00:42.953
In order for it to be a
calculus-based justification,

00:00:42.953 --> 00:00:44.500
we should use calculus in some way.

00:00:44.500 --> 00:00:47.310
So maybe use the derivative in some way.

00:00:47.310 --> 00:00:48.981
Now, you might recognize

00:00:48.981 --> 00:00:51.873
that you know that a
function is increasing

00:00:51.873 --> 00:00:54.361
if its derivative is positive.

00:00:54.361 --> 00:00:56.403
So before I even look at
what the students said,

00:00:56.403 --> 00:00:59.341
what I would say, my
calculus-based justification,

00:00:59.341 --> 00:01:01.544
and I wouldn't even have
to see the graph of h,

00:01:01.544 --> 00:01:03.418
I would just have to
see the graph of h prime

00:01:03.418 --> 00:01:07.166
is to say, look, h prime is greater than,

00:01:07.166 --> 00:01:10.214
h prime is positive when
x is greater than zero.

00:01:10.214 --> 00:01:12.332
If the derivative is positive

00:01:12.332 --> 00:01:13.272
then that means that the slope

00:01:13.272 --> 00:01:14.866
of the tangent line is positive

00:01:14.866 --> 00:01:17.635
and that means that the graph
of the original function

00:01:17.635 --> 00:01:18.980
is going to be increasing.

00:01:18.980 --> 00:01:20.366
Now, let's see whether one
of the students said that

00:01:20.366 --> 00:01:23.019
or what some of the other students wrote.

00:01:23.019 --> 00:01:25.707
So can you match the teacher's comments

00:01:25.707 --> 00:01:27.384
to the justifications?

00:01:27.384 --> 00:01:30.529
So one student wrote, the
derivative of h is increasing

00:01:30.529 --> 00:01:32.733
when x is greater than zero.

00:01:32.733 --> 00:01:36.373
So it is indeed the case that
the derivative is increasing

00:01:36.373 --> 00:01:38.462
when x is greater than zero

00:01:38.462 --> 00:01:41.488
but that's not the justification
for why h is increasing.

00:01:41.488 --> 00:01:43.605
For example, the derivative
could be increasing

00:01:43.605 --> 00:01:45.034
while still being negative

00:01:45.034 --> 00:01:47.563
in which case h would be decreasing.

00:01:47.563 --> 00:01:50.752
The appropriate justification
is that h prime is positive,

00:01:50.752 --> 00:01:52.218
not that it's necessarily increasing

00:01:52.218 --> 00:01:54.909
'cause you could be increasing
and still not be positive.

00:01:54.909 --> 00:01:55.971
So let's see.

00:01:55.971 --> 00:02:00.138
I would say that this doesn't
justify why h is increasing.

00:02:02.879 --> 00:02:05.044
When x is greater than zero,

00:02:05.044 --> 00:02:09.211
as the x-values increase, the
function values also increase.

00:02:10.100 --> 00:02:14.559
Well, that is a justification
for why h is increasing

00:02:14.559 --> 00:02:15.660
but that's not calculus-based.

00:02:15.660 --> 00:02:17.786
In no way are you using a derivative.

00:02:17.786 --> 00:02:21.824
So this isn't a
calculus-based justification.

00:02:21.824 --> 00:02:23.657
It's above the x-axis.

00:02:25.232 --> 00:02:27.232
So this one, what is it?

00:02:28.089 --> 00:02:29.137
Are they talking about h?

00:02:29.137 --> 00:02:31.460
Are they talking about h prime?

00:02:31.460 --> 00:02:35.800
If they were saying that h
prime is above the x-axis

00:02:35.800 --> 00:02:39.571
when x is greater than zero
then that would be a good answer

00:02:39.571 --> 00:02:42.834
but this is just what is above the x-axis

00:02:42.834 --> 00:02:44.345
and over what interval?

00:02:44.345 --> 00:02:46.874
So I would actually, let's
scroll down a little bit,

00:02:46.874 --> 00:02:49.405
this looks like a good thing
for the teacher to write.

00:02:49.405 --> 00:02:51.403
Please use more precise language.

00:02:51.403 --> 00:02:54.424
This cannot be accepted as
a correct justification.

00:02:54.424 --> 00:02:56.425
And then finally, this last student wrote,

00:02:56.425 --> 00:02:59.899
the derivative of h is positive
when x is greater than zero

00:02:59.899 --> 00:03:00.841
and it is indeed the case.

00:03:00.841 --> 00:03:03.546
If your derivative is positive,

00:03:03.546 --> 00:03:05.297
that means that your original function

00:03:05.297 --> 00:03:07.751
is going to be increasing
over that interval.

00:03:07.751 --> 00:03:09.918
So kudos, you are correct.


Conditions for MVT: table

WEBVTT
Kind: captions
Language: en

00:00:00.822 --> 00:00:03.353
- [Instructor] So we've been
given the value of h of x

00:00:03.353 --> 00:00:06.348
at a few values of x, and then we're told,

00:00:06.348 --> 00:00:10.110
James said that since h
of seven minus h of three

00:00:10.110 --> 00:00:13.825
over seven minus three is equal to one.

00:00:13.825 --> 00:00:16.797
So this is really the
average rate of change

00:00:16.797 --> 00:00:19.885
between x is equal to three
and x is equal to seven,

00:00:19.885 --> 00:00:22.741
between that point and that
point right over there.

00:00:22.741 --> 00:00:24.297
So since that is equal to one,

00:00:24.297 --> 00:00:26.921
there must be a number
c in the closed interval

00:00:26.921 --> 00:00:29.638
from three to seven for
which the derivative

00:00:29.638 --> 00:00:33.005
at that value c is equal to one.

00:00:33.005 --> 00:00:36.046
So what James is trying to do is apply

00:00:36.046 --> 00:00:38.763
the Mean Value Theorem,

00:00:38.763 --> 00:00:41.642
which tells us if the conditions apply

00:00:41.642 --> 00:00:43.616
for the Mean Value Theorem,

00:00:43.616 --> 00:00:47.699
it tells us that if I'm
going through two values,

00:00:49.259 --> 00:00:52.463
so let's say this is
a, let's say this is b,

00:00:52.463 --> 00:00:55.435
and let's say the function
does something like this.

00:00:55.435 --> 00:00:57.943
If we meet the conditions
for the Mean Value Theorem,

00:00:57.943 --> 00:01:00.915
it tells us that there's
some c in the closed interval

00:01:00.915 --> 00:01:04.677
from a to b where the
derivative of c is equal to

00:01:04.677 --> 00:01:07.347
the average rate of
change between a and b.

00:01:07.347 --> 00:01:09.344
So the average rate of
change between a and b

00:01:09.344 --> 00:01:13.511
would be the slope of the
secant line right over there,

00:01:14.731 --> 00:01:17.285
and then we could just think about well,

00:01:17.285 --> 00:01:19.328
looks like there are some points.

00:01:19.328 --> 00:01:20.420
The way I've drawn it,

00:01:20.420 --> 00:01:22.509
that point right over there
seems to have the same slope.

00:01:22.509 --> 00:01:26.318
This point right over there
seems to have the same slope.

00:01:26.318 --> 00:01:28.570
And so that's all what the
Mean Value Theorem is claiming,

00:01:28.570 --> 00:01:30.427
that there's going to be at least one c,

00:01:30.427 --> 00:01:33.005
if we meet the conditions
for the Mean Value Theorem,

00:01:33.005 --> 00:01:35.466
where the derivative at
that point is the same

00:01:35.466 --> 00:01:38.601
as the average rate of change
from the first endpoint

00:01:38.601 --> 00:01:39.808
to the second.

00:01:39.808 --> 00:01:42.432
Now what are the conditions
for the Mean Value Theorem

00:01:42.432 --> 00:01:45.404
to apply, and we've reviewed
this in multiple videos.

00:01:45.404 --> 00:01:46.728
One way to think about it,

00:01:46.728 --> 00:01:48.191
if we're talking about the closed interval

00:01:48.191 --> 00:01:49.677
from three to seven,

00:01:49.677 --> 00:01:53.462
one condition is that you
have to be differentiable,

00:01:53.462 --> 00:01:56.629
differentiable, over the open interval

00:01:58.152 --> 00:01:59.592
from three to seven.

00:01:59.592 --> 00:02:02.378
So that's the interval but
not including the endpoints.

00:02:02.378 --> 00:02:05.443
And you have to be continuous

00:02:05.443 --> 00:02:09.550
continuous over the entire closed interval

00:02:09.550 --> 00:02:11.431
so including the endpoints.

00:02:11.431 --> 00:02:13.335
And one interesting thing
that we've mentioned before

00:02:13.335 --> 00:02:15.912
is that differentiability
implies continuity.

00:02:15.912 --> 00:02:18.188
So if something is differentiable
over the open interval,

00:02:18.188 --> 00:02:21.648
it's also going to be continuous
over this open interval,

00:02:21.648 --> 00:02:24.132
and so the second condition
would just say well then we also

00:02:24.132 --> 00:02:26.965
have to be continuous at the endpoints.

00:02:26.965 --> 00:02:28.614
Now let's look at the answers.

00:02:28.614 --> 00:02:31.493
So it says which condition
makes James's claim true?

00:02:31.493 --> 00:02:33.838
So we have to feel good
about these two things

00:02:33.838 --> 00:02:36.857
right over here in order
to make James's claim.

00:02:36.857 --> 00:02:41.036
Choice A, h is continuous
over the closed interval

00:02:41.036 --> 00:02:43.126
from three to seven.

00:02:43.126 --> 00:02:46.145
So that does meet this second condition,

00:02:46.145 --> 00:02:49.094
but continuity does not
imply differentiability.

00:02:49.094 --> 00:02:50.510
So that doesn't give us the confidence

00:02:50.510 --> 00:02:52.971
that we're differentiable
over the open interval.

00:02:52.971 --> 00:02:54.527
If you're differentiable
you're continuous,

00:02:54.527 --> 00:02:55.758
but if you're continuous

00:02:55.758 --> 00:02:57.546
you're not necessarily differentiable.

00:02:57.546 --> 00:03:00.750
Classic example of that is
if we have a sharp turn,

00:03:00.750 --> 00:03:01.888
something like that,

00:03:01.888 --> 00:03:03.560
we wouldn't be
differentiable at that point

00:03:03.560 --> 00:03:06.207
even though we are continuous there.

00:03:06.207 --> 00:03:08.389
So let me rule that one out.

00:03:08.389 --> 00:03:10.224
I'm gonna rule that one out.

00:03:10.224 --> 00:03:13.557
Choice B, h is continuous and decreasing

00:03:14.427 --> 00:03:18.177
over the closed interval
from three to seven.

00:03:19.489 --> 00:03:21.021
Now that doesn't help us either

00:03:21.021 --> 00:03:23.482
because it still doesn't
mean you're differentiable.

00:03:23.482 --> 00:03:26.501
You could be continuous
and then decreasing

00:03:26.501 --> 00:03:27.894
and still have one of these sharp turns

00:03:27.894 --> 00:03:29.705
where you're not differentiable.

00:03:29.705 --> 00:03:32.910
So we will rule this one out.

00:03:32.910 --> 00:03:35.255
H is differentiable
over the closed interval

00:03:35.255 --> 00:03:37.043
from three to seven.

00:03:37.043 --> 00:03:38.273
This one feels good

00:03:38.273 --> 00:03:40.572
because if you're differentiable
over the closed interval,

00:03:40.572 --> 00:03:42.082
you're definitely gonna be differentiable

00:03:42.082 --> 00:03:44.868
over the open interval that
does not include the endpoints.

00:03:44.868 --> 00:03:47.422
This is a subset of this right over here.

00:03:47.422 --> 00:03:49.303
And if you're differentiable
over a closed interval,

00:03:49.303 --> 00:03:50.882
you're gonna be continuous over it.

00:03:50.882 --> 00:03:53.227
Differentiability implies continuity.

00:03:53.227 --> 00:03:56.501
So I like this choice right over here.

00:03:56.501 --> 00:04:00.216
Now this last choice says the
limit as x approaches five

00:04:00.216 --> 00:04:03.133
of h prime of x is equal to one.

00:04:05.696 --> 00:04:08.436
So they're saying the
limit of our derivative

00:04:08.436 --> 00:04:11.436
as we approach five is equal to one.

00:04:14.172 --> 00:04:16.052
Now that doesn't, we don't know for sure,

00:04:16.052 --> 00:04:17.515
this limit might be true,

00:04:17.515 --> 00:04:19.350
but that still does not necessarily imply

00:04:19.350 --> 00:04:22.554
that h prime of five is equal to one.

00:04:22.554 --> 00:04:24.551
We sill don't know that.

00:04:24.551 --> 00:04:26.107
And you know five is in this interval,

00:04:26.107 --> 00:04:28.823
but we still don't know just
from this statement alone

00:04:28.823 --> 00:04:31.912
that there is definitely
some c in the interval

00:04:31.912 --> 00:04:34.907
whose derivative is the same
as the average rate of change

00:04:34.907 --> 00:04:36.741
over the interval.

00:04:36.741 --> 00:04:39.824
So I would rule this one out as well.


Conditions for MVT: graph

WEBVTT
Kind: captions
Language: en

00:00:00.682 --> 00:00:01.515
- [Instructor] So we're asked,

00:00:01.515 --> 00:00:05.538
does the mean value theorem
apply to h over the interval?

00:00:05.538 --> 00:00:08.090
And they actually give us
four different intervals here.

00:00:08.090 --> 00:00:09.634
So we should separately consider them.

00:00:09.634 --> 00:00:11.967
And this is the graph
of y is equal to h of x.

00:00:11.967 --> 00:00:13.268
So pause this video and see,

00:00:13.268 --> 00:00:15.288
does the mean value theorem apply to h

00:00:15.288 --> 00:00:18.871
over any or all or some
of these intervals?

00:00:19.908 --> 00:00:21.729
All right, now let's work
through this together.

00:00:21.729 --> 00:00:24.320
In other videos, if the
words mean value theorem

00:00:24.320 --> 00:00:26.271
are completely unfamiliar to you,

00:00:26.271 --> 00:00:28.302
I encourage you to look
at the mean value theorem

00:00:28.302 --> 00:00:30.578
introduction, or the existence
theorem introductions

00:00:30.578 --> 00:00:31.890
on Khan Academy.

00:00:31.890 --> 00:00:33.434
But as a review for those of you

00:00:33.434 --> 00:00:35.686
who are familiar with
the mean value theorem,

00:00:35.686 --> 00:00:37.126
the conditions are,

00:00:37.126 --> 00:00:39.564
if we're dealing with some closed interval

00:00:39.564 --> 00:00:40.564
from a to b,

00:00:41.526 --> 00:00:44.034
we have to be differentiable,

00:00:44.034 --> 00:00:45.474
all right, just diff,

00:00:45.474 --> 00:00:46.945
actually, I'll just write it out,

00:00:46.945 --> 00:00:48.112
differentiable

00:00:50.625 --> 00:00:54.294
over the open interval from a to b,

00:00:54.294 --> 00:00:56.627
and we have to be continuous

00:00:58.381 --> 00:01:01.464
over the closed interval from a to b.

00:01:03.269 --> 00:01:04.151
Or another way to think about it,

00:01:04.151 --> 00:01:05.045
if you're differentiable,

00:01:05.045 --> 00:01:06.903
you're definitely going to be continuous.

00:01:06.903 --> 00:01:09.747
So, this second condition,

00:01:09.747 --> 00:01:11.465
just make sure that we're continuous

00:01:11.465 --> 00:01:14.578
at the endpoints of our closed interval.

00:01:14.578 --> 00:01:17.219
But if these conditions are met,

00:01:17.219 --> 00:01:19.377
then the mean value theorem tells us,

00:01:19.377 --> 00:01:20.479
and I'll just visually draw it

00:01:20.479 --> 00:01:22.245
because this is a review here.

00:01:22.245 --> 00:01:25.495
So let's say that this is our function.

00:01:27.090 --> 00:01:29.156
So let's say our function looks like this,

00:01:29.156 --> 00:01:31.616
this is a and this is b,

00:01:31.616 --> 00:01:33.322
and let's say that we
meet these conditions

00:01:33.322 --> 00:01:34.454
over this interval,

00:01:34.454 --> 00:01:37.095
the mean value theorem tells us

00:01:37.095 --> 00:01:39.993
that there's going to be some value,

00:01:39.993 --> 00:01:41.849
let's call it c, in the interval,

00:01:41.849 --> 00:01:46.015
where the derivative of c is equal

00:01:46.015 --> 00:01:49.064
to the average rate of change from a to b.

00:01:49.064 --> 00:01:50.769
So the slope of the secant line

00:01:50.769 --> 00:01:53.018
is the average rate of change.

00:01:53.018 --> 00:01:54.286
And so you can see it visually,

00:01:54.286 --> 00:01:55.931
it looks like right about there,

00:01:55.931 --> 00:01:58.421
the slope of the tangent
line would be the same.

00:01:58.421 --> 00:02:01.108
And so this would be the c that exists.

00:02:01.108 --> 00:02:04.036
Now, in this video, we're not
trying to identify the c's,

00:02:04.036 --> 00:02:04.987
we're just trying to say,

00:02:04.987 --> 00:02:07.130
can we apply the mean value theorem?

00:02:07.130 --> 00:02:08.368
So let's look at this first interval

00:02:08.368 --> 00:02:11.031
from negative five to negative
one, a closed interval.

00:02:11.031 --> 00:02:15.257
So we're going from negative
five to negative one.

00:02:15.257 --> 00:02:18.895
So are we differentiable
over the interval,

00:02:18.895 --> 00:02:20.615
over the open interval?

00:02:20.615 --> 00:02:23.483
We have this discontinuity over here

00:02:23.483 --> 00:02:25.264
that's not going to
make it differentiable,

00:02:25.264 --> 00:02:26.546
and, of course, this is also,

00:02:26.546 --> 00:02:28.297
if we're not continuous, we're not going

00:02:28.297 --> 00:02:30.108
to be differentiable here.

00:02:30.108 --> 00:02:32.236
So this discontinuity here

00:02:32.236 --> 00:02:34.772
actually violates both
of these conditions.

00:02:34.772 --> 00:02:37.368
And so for this first
interval, we would say, no,

00:02:37.368 --> 00:02:40.733
the mean value theorem does not apply.

00:02:40.733 --> 00:02:44.024
Now, the second one, from
negative one to three.

00:02:44.024 --> 00:02:45.774
So I'll do this in a
slightly different color.

00:02:45.774 --> 00:02:48.778
So we're gonna go from
negative one, right over there,

00:02:48.778 --> 00:02:49.611
to three.

00:02:50.514 --> 00:02:55.072
And if you look over this
interval, it looks like,

00:02:55.072 --> 00:02:57.109
over that interval, over
that closed interval,

00:02:57.109 --> 00:02:59.967
our function just really
looks like a line.

00:02:59.967 --> 00:03:02.729
And so it is both continuous
and differentiable

00:03:02.729 --> 00:03:04.012
over that interval,

00:03:04.012 --> 00:03:05.944
and it makes sense that the
mean value theorem applies.

00:03:05.944 --> 00:03:09.189
Actually, every c on this interval is the

00:03:09.189 --> 00:03:11.317
derivative, is the
instantaneous rate of change

00:03:11.317 --> 00:03:12.773
equal to the average rate of change

00:03:12.773 --> 00:03:15.233
because it looks linear
over this interval.

00:03:15.233 --> 00:03:18.644
So the mean value theorem
definitely applies over there.

00:03:18.644 --> 00:03:21.708
Now, what about from three to seven?

00:03:21.708 --> 00:03:25.041
The closed interval from three to seven.

00:03:26.070 --> 00:03:28.168
So when we look at this closed interval,

00:03:28.168 --> 00:03:31.821
we're definitely continuous
over the interval.

00:03:31.821 --> 00:03:35.684
So we meet the second criteria,
but are we differentiable?

00:03:35.684 --> 00:03:38.099
Well, a good giveaway of
a point that is continuous

00:03:38.099 --> 00:03:40.288
but not differentiable is right over here.

00:03:40.288 --> 00:03:42.280
Whenever you have these sharp edges,

00:03:42.280 --> 00:03:44.755
because we don't have a well-defined

00:03:44.755 --> 00:03:47.660
slope of a tangent line here.

00:03:47.660 --> 00:03:50.966
Would it be that, would it
be that, would it be that?

00:03:50.966 --> 00:03:53.622
And so because we aren't
differentiable at that point,

00:03:53.622 --> 00:03:55.222
at x is equal to six,

00:03:55.222 --> 00:03:57.365
we aren't differentiable
over this open interval

00:03:57.365 --> 00:03:59.207
from three to seven,

00:03:59.207 --> 00:04:03.236
and so the mean value
theorem does not apply.

00:04:03.236 --> 00:04:04.897
And you can even see that visually.

00:04:04.897 --> 00:04:08.293
The average rate of change
between the endpoints

00:04:08.293 --> 00:04:10.270
of our interval,

00:04:10.270 --> 00:04:12.820
or if we want to think about
the slope of the secant line,

00:04:12.820 --> 00:04:14.345
that's that right over there,

00:04:14.345 --> 00:04:17.711
and over the interval,
we don't see any point

00:04:17.711 --> 00:04:19.257
where the instantaneous rate of change,

00:04:19.257 --> 00:04:20.827
where the slope of the tangent line

00:04:20.827 --> 00:04:23.800
is equal to the slope of the secant line

00:04:23.800 --> 00:04:27.015
between the endpoints of our interval.

00:04:27.015 --> 00:04:30.215
So once again, mean value
theorem would not apply.

00:04:30.215 --> 00:04:33.430
Now, what about from
negative three to two?

00:04:33.430 --> 00:04:34.486
I'll do this in orange.

00:04:34.486 --> 00:04:38.569
So from negative three,
right over there, to two.

00:04:40.297 --> 00:04:41.972
Well, if we look at the open interval

00:04:41.972 --> 00:04:43.828
from negative three to two,

00:04:43.828 --> 00:04:45.262
we're not considering what happens

00:04:45.262 --> 00:04:47.526
at negative three and at two,

00:04:47.526 --> 00:04:49.624
over the open interval,

00:04:49.624 --> 00:04:53.352
it does look like we are both
differentiable and continuous.

00:04:53.352 --> 00:04:55.948
So we're definitely differentiable
over the open interval.

00:04:55.948 --> 00:04:59.238
But clearly, we're not continuous
over the closed interval.

00:04:59.238 --> 00:05:03.313
At negative three, we
actually are not continuous.

00:05:03.313 --> 00:05:04.581
And so because of that,

00:05:04.581 --> 00:05:07.313
the mean value theorem does not apply.

00:05:07.313 --> 00:05:09.796
And you can actually even
see that it would not apply

00:05:09.796 --> 00:05:12.497
because if you look at the
slope of the secant line

00:05:12.497 --> 00:05:14.158
between our endpoints,

00:05:14.158 --> 00:05:16.361
that's the secant line right over there,

00:05:16.361 --> 00:05:20.497
and so you could see, over
that interval, there is no c

00:05:20.497 --> 00:05:22.232
between negative three and two,

00:05:22.232 --> 00:05:23.938
where the slope of the tangent line

00:05:23.938 --> 00:05:26.292
or the instantaneous
slope or the derivative

00:05:26.292 --> 00:05:30.171
is going to be the same as
the slope of the secant line.

00:05:30.171 --> 00:05:32.073
And that doesn't violate
the mean value theorem

00:05:32.073 --> 00:05:34.208
because the mean value theorem
just doesn't apply here,

00:05:34.208 --> 00:05:35.703
we haven't met the condition

00:05:35.703 --> 00:05:38.961
of being continuous over
the closed interval.


Introduction to integral calculus

WEBVTT
Kind: captions
Language: en

00:00:00.486 --> 00:00:01.691
- [Instructor] So I have a curve here

00:00:01.691 --> 00:00:05.284
that represents y is equal to f of x,

00:00:05.284 --> 00:00:06.682
and there's a classic problem

00:00:06.682 --> 00:00:09.394
that mathematicians
have long thought about.

00:00:09.394 --> 00:00:12.492
How do we find the area under this curve?

00:00:12.492 --> 00:00:15.372
Maybe under the curve
and above the x-axis,

00:00:15.372 --> 00:00:16.837
and let's say between two boundaries.

00:00:16.837 --> 00:00:21.251
Let's say between x is equal
to a and x is equal to b.

00:00:21.251 --> 00:00:23.246
So let me draw these
boundaries right over here.

00:00:23.246 --> 00:00:24.526
That's our left boundary.

00:00:24.526 --> 00:00:27.342
This is our right boundary.

00:00:27.342 --> 00:00:31.538
And we want to think about
this area right over here.

00:00:31.538 --> 00:00:33.793
Well, without calculus, you could actually

00:00:33.793 --> 00:00:36.298
get better and better
approximations for it.

00:00:36.298 --> 00:00:37.639
How would you do it?

00:00:37.639 --> 00:00:39.702
Well, you could divide this section

00:00:39.702 --> 00:00:42.839
into a bunch of delta
x's that go from a to b.

00:00:42.839 --> 00:00:44.603
They could be equal sections or not,

00:00:44.603 --> 00:00:46.748
but let's just say, for
the sake of visualizations,

00:00:46.748 --> 00:00:48.640
I'm gonna draw roughly
equal sections here.

00:00:48.640 --> 00:00:50.228
So that's the first.

00:00:50.228 --> 00:00:52.494
That's the second.

00:00:52.494 --> 00:00:54.320
This is the third.

00:00:54.320 --> 00:00:56.134
This is the fourth.

00:00:56.134 --> 00:00:57.699
This is the fifth.

00:00:57.699 --> 00:00:59.861
And then we have the
sixth right over here.

00:00:59.861 --> 00:01:02.944
And so each of these, this is delta x,

00:01:02.944 --> 00:01:04.992
let's just call that delta x one.

00:01:04.992 --> 00:01:06.813
This is delta x two.

00:01:06.813 --> 00:01:10.567
This width right over here,
this is delta x three,

00:01:10.567 --> 00:01:12.900
all the way to delta x n.

00:01:14.057 --> 00:01:15.744
I'll try to be general here.

00:01:15.744 --> 00:01:18.216
And so what we could do is,
let's try to sum up the area

00:01:18.216 --> 00:01:19.898
of the rectangles defined here.

00:01:19.898 --> 00:01:21.286
And we could make the height,

00:01:21.286 --> 00:01:23.838
maybe we make the height

00:01:23.838 --> 00:01:25.961
based on the value of the
function at the right bound.

00:01:25.961 --> 00:01:26.794
It doesn't have to be.

00:01:26.794 --> 00:01:28.518
It could be the value of the function

00:01:28.518 --> 00:01:30.503
someplace in this delta x.

00:01:30.503 --> 00:01:31.419
But that's one solution.

00:01:31.419 --> 00:01:33.068
We're gonna go into a
lot more depth into it

00:01:33.068 --> 00:01:34.522
in future videos.

00:01:34.522 --> 00:01:35.978
And so

00:01:35.978 --> 00:01:37.414
we do that.

00:01:37.414 --> 00:01:40.231
And so now we have an
approximation, where we could say,

00:01:40.231 --> 00:01:42.528
look, the area of each of these rectangles

00:01:42.528 --> 00:01:45.645
are going to be f of x sub i,

00:01:45.645 --> 00:01:47.963
where maybe x sub i is the right boundary,

00:01:47.963 --> 00:01:51.463
the way I've drawn it, times delta x i.

00:01:52.840 --> 00:01:54.535
That's each of these rectangles.

00:01:54.535 --> 00:01:57.206
And then we can sum them up,

00:01:57.206 --> 00:02:01.623
and that would give us an
approximation for the area.

00:02:01.623 --> 00:02:03.929
But as long as we use a finite number,

00:02:03.929 --> 00:02:06.145
we might say, well, we
can always get better by

00:02:06.145 --> 00:02:08.124
making our delta x's smaller

00:02:08.124 --> 00:02:12.211
and then by having more
of these rectangles,

00:02:12.211 --> 00:02:15.131
or get to a situation
here we're going from

00:02:15.131 --> 00:02:16.830
i is equal to one

00:02:16.830 --> 00:02:19.240
to i is equal to n.

00:02:19.240 --> 00:02:20.962
But what happens is delta x

00:02:20.962 --> 00:02:23.428
gets thinner and thinner and thinner,

00:02:23.428 --> 00:02:25.575
and n gets larger and larger and larger,

00:02:25.575 --> 00:02:28.733
as delta x gets infinitesimally small

00:02:28.733 --> 00:02:31.719
and then as n approaches infinity.

00:02:31.719 --> 00:02:33.778
And so you're probably sensing something,

00:02:33.778 --> 00:02:36.858
that maybe we could think about the limit

00:02:36.858 --> 00:02:39.844
as we could say as n approaches infinity

00:02:39.844 --> 00:02:41.809
or the limit as delta x

00:02:41.809 --> 00:02:44.269
becomes very, very, very, very small.

00:02:44.269 --> 00:02:47.603
And this notion of getting
better and better approximations

00:02:47.603 --> 00:02:50.531
as we take the limit as
n approaches infinity,

00:02:50.531 --> 00:02:54.114
this is the core idea
of integral calculus.

00:02:57.253 --> 00:02:59.743
And it's called integral calculus

00:02:59.743 --> 00:03:03.189
because the central operation we use,

00:03:03.189 --> 00:03:05.517
the summing up of an infinite number

00:03:05.517 --> 00:03:08.978
of infinitesimally thin things
is one way to visualize it,

00:03:08.978 --> 00:03:10.253
is the integral,

00:03:10.253 --> 00:03:12.242
that this is going to be the integral,

00:03:12.242 --> 00:03:14.686
in this case, from a to b.

00:03:14.686 --> 00:03:15.983
And we're gonna learn in a lot more depth,

00:03:15.983 --> 00:03:20.150
in this case, it is a
definite integral of f of x,

00:03:21.163 --> 00:03:22.080
f of x, dx.

00:03:23.075 --> 00:03:25.030
But you can already
see the parallels here.

00:03:25.030 --> 00:03:28.374
You can view the integral
sign as like a sigma notation,

00:03:28.374 --> 00:03:30.679
as a summation sign, but
instead of taking the sum

00:03:30.679 --> 00:03:32.196
of a discrete number of things

00:03:32.196 --> 00:03:34.544
you're taking the sum of an infinitely,

00:03:34.544 --> 00:03:37.907
an infinite number,
infinitely thin things.

00:03:37.907 --> 00:03:40.691
Instead of delta x, you now have dx,

00:03:40.691 --> 00:03:43.600
infinitesimally small things.

00:03:43.600 --> 00:03:45.454
And this is a notion of an integral.

00:03:45.454 --> 00:03:48.177
So this right over here is an integral.

00:03:48.177 --> 00:03:50.141
Now what makes it interesting to calculus,

00:03:50.141 --> 00:03:52.039
it is using this notion of a limit,

00:03:52.039 --> 00:03:54.087
but what makes it even more powerful

00:03:54.087 --> 00:03:56.602
is it's connected to the
notion of a derivative,

00:03:56.602 --> 00:03:59.449
which is one of these beautiful
things in mathematics.

00:03:59.449 --> 00:04:02.747
As we will see in the
fundamental theorem of calculus,

00:04:02.747 --> 00:04:05.194
that integration, the
notion of an integral,

00:04:05.194 --> 00:04:08.681
is closely, tied closely to
the notion of a derivative,

00:04:08.681 --> 00:04:11.076
in fact, the notion of an antiderivative.

00:04:11.076 --> 00:04:14.100
In differential calculus,
we looked at the problem of,

00:04:14.100 --> 00:04:17.407
hey, if I have some function,
I can take its derivative,

00:04:17.407 --> 00:04:20.160
and I can get the
derivative of the function.

00:04:20.160 --> 00:04:23.189
Integral calculus, we're
going to be doing a lot of,

00:04:23.189 --> 00:04:25.327
well, what if we start
with the derivative,

00:04:25.327 --> 00:04:26.744
can we figure out

00:04:27.648 --> 00:04:29.909
through integration,

00:04:29.909 --> 00:04:32.322
can we figure out its antiderivative

00:04:32.322 --> 00:04:34.917
or the function whose derivative it is?

00:04:34.917 --> 00:04:37.420
As we will see, all of these are related.

00:04:37.420 --> 00:04:39.874
The idea of the area under a curve,

00:04:39.874 --> 00:04:41.041
the idea of a limit

00:04:41.041 --> 00:04:43.500
of summing an infinite
number of infinitely things,

00:04:43.500 --> 00:04:46.827
thin things, and the notion
of an antiderivative,

00:04:46.827 --> 00:04:50.994
they all come together in our
journey in integral calculus.


Connecting f, f', and f'' graphically

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:02.607
- [Instructor] We have the
graphs of three functions here,

00:00:02.607 --> 00:00:05.299
and what we know is that one
of them is the function f,

00:00:05.299 --> 00:00:07.531
another is the first derivative of f,

00:00:07.531 --> 00:00:10.291
and then the third is the
second derivative of f.

00:00:10.291 --> 00:00:13.280
And our goal is to figure
out which function is which.

00:00:13.280 --> 00:00:15.004
Which one is f, which
is the first derivative,

00:00:15.004 --> 00:00:16.564
and which is the second?

00:00:16.564 --> 00:00:18.681
Like always, pause this video and see

00:00:18.681 --> 00:00:20.292
if you can work through it on your own

00:00:20.292 --> 00:00:22.424
before we do it together.

00:00:22.424 --> 00:00:24.428
All right, now let's do this together.

00:00:24.428 --> 00:00:26.870
The way I'm going to tackle
it is I'm gonna try to sketch

00:00:26.870 --> 00:00:30.971
what we can about the derivatives
of each of these graphs,

00:00:30.971 --> 00:00:34.376
or the functions
represented by these graphs.

00:00:34.376 --> 00:00:37.907
So in this first graph
here in this orange color,

00:00:37.907 --> 00:00:41.347
we can see that the slope
is quite positive here,

00:00:41.347 --> 00:00:44.196
but then it becomes less
and less and less positive,

00:00:44.196 --> 00:00:47.236
up until this point where the
slope is going to be zero,

00:00:47.236 --> 00:00:49.344
and then it becomes more and more

00:00:49.344 --> 00:00:51.243
and more and more negative.

00:00:51.243 --> 00:00:54.930
So the derivative of this
curve right over here,

00:00:54.930 --> 00:00:57.391
or the function represented by this curve

00:00:57.391 --> 00:01:01.457
is gonna start off reasonably
positive right over there.

00:01:01.457 --> 00:01:03.919
At this point, the derivative
is gonna cross zero,

00:01:03.919 --> 00:01:05.321
'cause our derivative is zero there,

00:01:05.321 --> 00:01:06.797
slope of the tangent line.

00:01:06.797 --> 00:01:08.655
And then it's gonna get
more and more negative,

00:01:08.655 --> 00:01:10.967
at least over the interval that we see.

00:01:10.967 --> 00:01:13.733
So it might look, I don't
know, something like this.

00:01:13.733 --> 00:01:15.002
I don't know if it's a line or not.

00:01:15.002 --> 00:01:17.507
It might be some type of a curve.

00:01:17.507 --> 00:01:20.735
It would definitely have a
trend something like that.

00:01:20.735 --> 00:01:21.828
Now, we could immediately tell

00:01:21.828 --> 00:01:23.890
that this blue graph is not the derivative

00:01:23.890 --> 00:01:25.296
of this orange graph.

00:01:25.296 --> 00:01:26.450
Its trend is opposite.

00:01:26.450 --> 00:01:27.429
Over that interval,

00:01:27.429 --> 00:01:29.808
it's going from being
negative to positive,

00:01:29.808 --> 00:01:32.847
as opposed to going from
positive to negative.

00:01:32.847 --> 00:01:34.542
So we can rule out the blue graph

00:01:34.542 --> 00:01:37.441
as being the derivative
of the orange graph.

00:01:37.441 --> 00:01:39.184
But what about this magenta graph?

00:01:39.184 --> 00:01:42.839
It does look like it has the right trend.

00:01:42.839 --> 00:01:46.345
In fact, it intersects the
x-axis at the right place

00:01:46.345 --> 00:01:47.850
right over there.

00:01:47.850 --> 00:01:49.517
And at least over this interval,

00:01:49.517 --> 00:01:52.715
it seems it's positive from here to here.

00:01:52.715 --> 00:01:54.669
So it's positive.

00:01:54.669 --> 00:01:56.155
This graph is positive

00:01:56.155 --> 00:01:58.876
when the slope of the tangent
line here is positive.

00:01:58.876 --> 00:02:00.687
And this graph is negative

00:02:00.687 --> 00:02:04.175
when the slope of the tangent
line here is negative.

00:02:04.175 --> 00:02:06.538
Now, one thing that might
be causing some unease

00:02:06.538 --> 00:02:09.676
to immediately say that this last graph

00:02:09.676 --> 00:02:11.580
is the derivative of the first one

00:02:11.580 --> 00:02:13.373
is we're not used to situations

00:02:13.373 --> 00:02:17.082
where the derivative
has more extreme points,

00:02:17.082 --> 00:02:20.688
more minima and maxima
than the original function.

00:02:20.688 --> 00:02:22.440
But in this case, it could just be

00:02:22.440 --> 00:02:25.135
'cause we don't see the
entire original function.

00:02:25.135 --> 00:02:28.603
So, for example, if this last
graph is indeed the derivative

00:02:28.603 --> 00:02:30.443
of this first graph,

00:02:30.443 --> 00:02:34.276
then what we see is our
derivative is negative

00:02:35.746 --> 00:02:36.579
right over here,

00:02:36.579 --> 00:02:39.432
but then right around here it
starts becoming less negative.

00:02:39.432 --> 00:02:42.940
So if that point corresponds
to roughly right over there,

00:02:42.940 --> 00:02:47.569
then over here our slope
will become less and less

00:02:47.569 --> 00:02:48.922
and less negative,

00:02:48.922 --> 00:02:52.371
and then at this point our
slope would become zero,

00:02:52.371 --> 00:02:54.325
which would be right around there.

00:02:54.325 --> 00:02:59.253
So for example, our graph
might look something like this,

00:02:59.253 --> 00:03:00.172
we just didn't see it.

00:03:00.172 --> 00:03:02.511
It fell off of the part of the graph

00:03:02.511 --> 00:03:04.034
that we actually showed.

00:03:04.034 --> 00:03:05.055
So I would actually say

00:03:05.055 --> 00:03:07.286
that this is a good candidate for being,

00:03:07.286 --> 00:03:08.864
the third function is a good candidate

00:03:08.864 --> 00:03:10.899
for being the derivative
of the first function.

00:03:10.899 --> 00:03:13.208
So maybe we could say that this is f

00:03:13.208 --> 00:03:15.862
and that this is f prime.

00:03:15.862 --> 00:03:17.564
Now, let's look at the second graph.

00:03:17.564 --> 00:03:19.793
What would its derivative look like?

00:03:19.793 --> 00:03:22.797
So over here our slope is quite negative,

00:03:22.797 --> 00:03:25.743
and it becomes less and
less and less negative

00:03:25.743 --> 00:03:28.404
until we go right over here
where our slope is zero.

00:03:28.404 --> 00:03:30.706
So our derivative would
intersect the x-axis

00:03:30.706 --> 00:03:31.539
right over there.

00:03:31.539 --> 00:03:33.192
It would start out negative,

00:03:33.192 --> 00:03:36.735
and it would become less
and less and less negative.

00:03:36.735 --> 00:03:38.835
And at this point it crosses the x-axis

00:03:38.835 --> 00:03:42.591
and it becomes more and more positive.

00:03:42.591 --> 00:03:44.882
So we see here our derivative

00:03:44.882 --> 00:03:46.629
becomes more and more positive.

00:03:46.629 --> 00:03:47.973
But then right around here it seems

00:03:47.973 --> 00:03:50.021
like it's getting less positive again.

00:03:50.021 --> 00:03:53.021
So it might look something like this

00:03:54.134 --> 00:03:56.786
where over here it's
becoming less positive again,

00:03:56.786 --> 00:03:58.518
less positive, less
positive, less positive.

00:03:58.518 --> 00:04:01.812
Right over here, our
derivative would be zero.

00:04:01.812 --> 00:04:04.157
Our derivative would
intersect the x-axis there.

00:04:04.157 --> 00:04:06.010
And then it just looks like it is,

00:04:06.010 --> 00:04:08.396
the slope is getting more
and more and more negative,

00:04:08.396 --> 00:04:09.473
so our derivative is gonna get more

00:04:09.473 --> 00:04:10.949
and more and more negative.

00:04:10.949 --> 00:04:12.800
Well, what I very
roughly just sketched out

00:04:12.800 --> 00:04:16.413
looks an awful lot like the
brown graph right over here.

00:04:16.413 --> 00:04:18.168
So this brown graph does indeed look

00:04:18.168 --> 00:04:20.456
like the derivative of this blue graph.

00:04:20.456 --> 00:04:23.569
So what I would say is
that this is actually f,

00:04:23.569 --> 00:04:25.484
and then this would be f prime.

00:04:25.484 --> 00:04:26.710
And then if this is f prime,

00:04:26.710 --> 00:04:29.963
the derivative of that is
going to be f prime prime.

00:04:29.963 --> 00:04:31.266
So that looks good.

00:04:31.266 --> 00:04:32.697
I would actually go with this.

00:04:32.697 --> 00:04:34.309
And if you wanted, just for safe measure,

00:04:34.309 --> 00:04:35.417
you could try to sketch out

00:04:35.417 --> 00:04:37.442
what the derivative of
this graph would be.

00:04:37.442 --> 00:04:39.022
Actually, let's just do that.

00:04:39.022 --> 00:04:42.323
So over here, the derivative of this,

00:04:42.323 --> 00:04:44.665
so right now we have a positive slope

00:04:44.665 --> 00:04:47.574
of our tangent line is getting
less and less positive.

00:04:47.574 --> 00:04:50.198
It hits zero right over there.

00:04:50.198 --> 00:04:53.345
So the derivative might
look something like this

00:04:53.345 --> 00:04:54.368
over the interval.

00:04:54.368 --> 00:04:56.531
Now, the slope of the
tangent line is getting more

00:04:56.531 --> 00:04:58.821
and more and more and more negative,

00:04:58.821 --> 00:05:01.209
right until about that point.

00:05:01.209 --> 00:05:03.508
So it's getting more and
more and more negative

00:05:03.508 --> 00:05:04.341
until about that point.

00:05:04.341 --> 00:05:06.032
And now it looks like
it's getting less and less

00:05:06.032 --> 00:05:07.634
and less and less negative,

00:05:07.634 --> 00:05:11.954
all the way until the derivative
goes back to being zero.

00:05:11.954 --> 00:05:13.838
And then it looks like
it's getting more and more

00:05:13.838 --> 00:05:16.260
and more and more positive.

00:05:16.260 --> 00:05:18.705
So the derivative of
this magenta curve looks

00:05:18.705 --> 00:05:21.297
like an upward opening U.

00:05:21.297 --> 00:05:23.227
And we don't see that over here,

00:05:23.227 --> 00:05:24.627
so we could feel good

00:05:24.627 --> 00:05:26.720
that its derivative
actually isn't depicted.

00:05:26.720 --> 00:05:29.579
So I feel calling the middle graph f,

00:05:29.579 --> 00:05:31.556
calling the left graph f prime,

00:05:31.556 --> 00:05:34.897
and calling the right graph
the second derivative.