One-to-One Intervals Problem #5 - UGNotesOnline

[lutya2]

Homework Statement

I'm trying to do problem #5 on this worksheet
http://ugnotesonline.com/attachments/008_t1.5pst.pdf

Homework Equations

none

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.

 

[rasmhop]

(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.

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.