Q:

Rewrite formula for a rectangle under graph

Given the graph of $f(x) = \frac{1}{2x - 1}$, for every $p > \frac{1}{2}$, the rectangle $(0,0), (p, 0), (p, f(p)), (0, f(p))$ is denoted by $A(p)$. So $A(p) = \frac{p}{2p -1}$. For certain values of $a$ and $b$ the area can be written as $A(p) = a + bf(p)$, for all $p > \frac{1}{2}$. Find $a$ and $b$.
With a hint: 'write the right-hand side as a single fraction.
Since I have no other ideas, I start with that hint: $A(p) = a + bf(p) = \frac{a(2x - 1) + b}{2x - 1}$. I guess I know the rectangle has a constant size, no matter what $p > \frac{1}{2}$ I decide on.
But what to do next? The only thing I could think of was just taking some $p > \frac{1}{2}$, so: $A(p) = \frac{a}{x} + \frac{b}{x}$. This doesn't seem particulary helpful.

A:

It can help to sub in two values of p, 1 and 2 (although you can choose can be anything >1/2).
The area of the rectangle you can find by using (base)(height), or p(f(p)). So A(1)=1, and A(2)=2/3.
Then you have two simultaneous equations:
A(1)=a+b(f(1))=1, or a+b=1
A(2)=a+b(f(2))=2/3, or 3a+b=2
Then you can solve simultaneously to find a and b.

