# Thread: Difficult BC Composition Problem

1. ## Difficult BC Composition Problem

Let $\displaystyle f$ and $\displaystyle g$ be differentiable functions with the following properties:

I. f(x) < 0 for all x
II. g(5)=2

If $\displaystyle h(x) = f(x)/g(x)$ and $\displaystyle h'(x)=$ $\displaystyle f'(x)/g(x)$, then $\displaystyle g(x)$

I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

Thanks!

2. Originally Posted by r2d2
Let $\displaystyle f$ and $\displaystyle g$ be differentiable functions with the following properties:

I. f(x) < 0 for all x
II. g(5)=2

If $\displaystyle h(x) = f(x)/g(x)$ and $\displaystyle h'(x)=$ $\displaystyle f'(x)/g(x)$, then $\displaystyle g(x)$

I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

Thanks!
$\displaystyle h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}=\frac{f'(x)}{g(x)}$

And we can use property 2 to say

$\displaystyle h'(5)=\frac{2f'(x)-f(x)g'(x)}{4}=\frac{f'(x)}{2}$

So, $\displaystyle 4f'(x)-2f(x)g'(x)=4f'(x)$

So, $\displaystyle 2f(x)g'(x)=0$

But $\displaystyle f(x)<0$ so $\displaystyle f(x)\not =0$ and therefore
$\displaystyle g'(x)=0$

So $\displaystyle g(x)=2$

3. Originally Posted by r2d2
Let $\displaystyle f$ and $\displaystyle g$ be differentiable functions with the following properties:

I. f(x) < 0 for all x
II. g(5)=2

If $\displaystyle h(x) = f(x)/g(x)$ and $\displaystyle h'(x)=$ $\displaystyle f'(x)/g(x)$, then $\displaystyle g(x)$

I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

Thanks!
I don't really understand what you want to do... do you want to find $\displaystyle g(x)$ ?

If so:

$\displaystyle h'(x) = \frac{f'(x)}{g(x)} = \frac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2} \Rightarrow \frac{f'(x)}{g(x)} - \frac{f'(x)g(x)}{(g(x))^2} + \frac{f(x)g'(x)}{(g(x))^2}=0$

$\displaystyle \Rightarrow \frac{f'(x)}{g(x)} - \frac{f'(x)}{g(x)} + \frac{f(x)g'(x)}{(g(x))^2} = 0 \Rightarrow f(x)g'(x) = 0$

But we know that $\displaystyle f(x) < 0 \Rightarrow f(x) \neq 0$ and therefore $\displaystyle g'(x) = 0 \Rightarrow g(x) = C$ where C is a constant.

We know that $\displaystyle g(5) = 2 \Rightarrow g(x) = 2$

4. Thank you for borth of your responses. I think where I got confused was whether or not g(x) had to be a function or a number, so at first I disregarded the second information.

Looking at it again, this problem seems so simple if I think of the right processes.

Thanks again!

R2