# Homework Help: One-to-One Intervals

1. Dec 12, 2009

### lutya2

1. The problem statement, all variables and given/known data
I'm trying to do problem #5 on this worksheet
http://ugnotesonline.com/attachments/008_t1.5pst.pdf [Broken]

2. Relevant equations
none

3. The attempt at a solution
(a) I tried to take the derivitive of the function in the intergral so I could see where it was increasing and decreasing and prove that it was one-to-one but that didn't work.

(b) I don't even know what this part is trying to ask.

Last edited by a moderator: May 4, 2017
2. Dec 12, 2009

### rasmhop

Try again and show us your work because that is as far as I can see the easiest approach. Remember that if f is everywhere differentiable and the derivative is strictly positive, then the function is strictly increasing.

For b, the notation $f^{-1}(x)$ denotes the inverse function. That is the function such that
$$f^{-1}(f(x)) = f(f^{-1}(x)) = x$$
for all x. (its existence is guaranteed by part a). To solve this part use the chain rule as follows:
$$\frac{d(f \circ f^{-1})(x)}{df^{-1}} \times \frac{df^{-1}(x)}{dx} = \frac{d(f^{-1} \circ f)(x)}{dx}$$
You should be able to evaluate everything here except $$\frac{df^{-1}(x)}{dx}$$ so you can solve the problem.

HINT:
$$\frac{d(f \circ f^{-1})(x)}{df^{-1}} = \frac{df(x)}{dx}$$

EDIT:
Also in case you didn't know $$g|_{x=0}$$ just means g(0), so in b you're just asked to evaluate the derivate at 0.