## gabie1121 Group Title find the values for c such that the function is continuous at all x. FUNCTION BELOW one year ago one year ago

1. gabie1121 Group Title

$f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}$

2. gabie1121 Group Title

$f(x)=(e ^{x}-2c) if x \ge0$

3. gabie1121 Group Title

those go together. I just couldn't figure out how to get them all in at once

4. zepdrix Group Title

In order for this function to be continuous, we need to pick a $$c$$ value that makes the two pieces connect together nicely at $$x=0$$. Imagine railroad tracks, the track needs to be continuous, and can't have any sharp corners or the train will fly off the tracks.

5. gabie1121 Group Title

Yep, that i get. So do i set both functions equal to 0 and solve for c or something?

6. zepdrix Group Title

We want to look at them in the limit. We want to see what is happening when we get closer and closer to 0 from the left side. And also what is happening when we get closer and closer to 0 from the right side. For this function to be continuous, they must be approaching the same value. So we'll set these limits equal to one another.

7. zepdrix Group Title

$\large \lim_{x \rightarrow 0^-}f(x) \quad = \quad \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x}$

8. zepdrix Group Title

Understand why our function is sin(cx)/x when we're approaching from the left?

9. zepdrix Group Title

The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.

10. gabie1121 Group Title

yep cuz it says less than 0

11. zepdrix Group Title

Ok cool :) so we'll set that equal to the other piece.

12. zepdrix Group Title

And solve for c.

13. gabie1121 Group Title

do we make the other side a limit as well?

14. zepdrix Group Title

$\large \lim_{x \rightarrow 0^-}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)$$\large \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x-2c$

15. zepdrix Group Title

Yes :) the limits need to agree in order for this function to be continuous.

16. gabie1121 Group Title

alrighty, well the left side is equal to just c, right?

17. zepdrix Group Title

Yes very good ^^

18. gabie1121 Group Title

because if i multiply by C , i can use the rule that says sinx/x=1

19. zepdrix Group Title

you remembered your identity i take it hehe

20. gabie1121 Group Title

yep! the right side is giving me fits though lol

21. zepdrix Group Title

Or would it be -1 since we're coming from the left? Hmm I didn't think about that. lemme check real quick.

22. zepdrix Group Title

Nah it's still 1, my bad.

23. gabie1121 Group Title

well would the right just be e^x-2? Or can i not do that cuz i'm thinking of derivative rules?

24. gabie1121 Group Title

well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?

25. zepdrix Group Title

no we're not thinking of this as a derivative :) We're looking at the limit and saying to ourselves, "If I plug x=0 directly into this function, does it cause a problem?" If the answer is no, then we can do just that!

26. gabie1121 Group Title

oh! well then we're left with $c=2c$ ?

27. zepdrix Group Title

Woops! Recall that if we have a 0 in the exponent, what will that change our base to?

28. zepdrix Group Title

Not 0 silly! :O

29. gabie1121 Group Title

oh well i just plugged in e^0 in my calculator and it gave me 1

30. zepdrix Group Title

hah XD that's a way to do it i guess! :D

31. zepdrix Group Title

yah 1 :3

32. gabie1121 Group Title

lol so its actuallly c=1-2c?

33. zepdrix Group Title

Yah looks good c:

34. zepdrix Group Title

Do you by chance have an answer key that we can check this against? This is one of those annoying problems that it's easy to make a mistake on c: lol

35. gabie1121 Group Title

c=1/3! , and I don't have one yet. I will monday so its not a huge deal

36. gabie1121 Group Title

It looks logical enough to me. Thanks, AGAIN lol

37. zepdrix Group Title

This type of problem becomes a little bit harder when they throw 2 unknown constants at you. Because then you have to also look at the limits of their derivatives. But this was a good problem to get a feel for the concept c:

38. gabie1121 Group Title

well i just started calc1 this semester, so i haven't learned much yet. Just dipping my toes in the water. I have to take all the way through calculus 3 though, bleh

39. zepdrix Group Title

Hmm you'll do quite well, I can tell. You seem quite smart. You're very quick on remember how to do little steps. Calc 2 is a doozy!! Power Series made me want to rip my hair out! :O

40. gabie1121 Group Title

Yeah, i'm told i'll want to murder myself with Calc 2. Not looking forward to it, but thanks! That makes me feel a little better

41. gabie1121 Group Title

i might be hunting you down again! lol