# Inverse function problem

• Sep 11th 2009, 10:24 PM
xxlvh
Inverse function problem
You are given that the funcion $\displaystyle f(x) = \frac{e^{cos x}-1}{2+e^{cos x}}$ is one-to-one when restricted on $\displaystyle [0,\pi]$

Find the domain of the inverse function of f, $\displaystyle x \epsilon [0,\pi]$
and find a formula for the inverse function.

I am totally stumped on this one.
• Sep 12th 2009, 02:11 AM
tutor.mathitutor
sol1.pdf

solution is attached.

• Sep 12th 2009, 06:08 PM
xxlvh
The equation of the inverse looks correct, but when I checked graphically the domain is different. The approximate decimal values were -0.267 < x < 0.364
• Sep 12th 2009, 06:29 PM
mr fantastic
Quote:

Originally Posted by tutor.mathitutor

solution is attached.

It is correct that the inverse function is given by the rule $\displaystyle y = \cos^{-1} \left( \ln \left( \frac{1 + 2x}{1 - x}\right) \right)$. However, it is incorrect to say that the maximal domain of $\displaystyle y = \cos^{-1} \left( \ln \left( \frac{1 + 2x}{1 - x}\right) \right)$ is the same as the domain of the inverse function. The domain of the inverse function is in fact a subset of the maximal domain.

@OP: To find the domain of the inverse function you should recall that $\displaystyle \text{dom} f^{-1} = \text{ran} f$. So your job boils down to finding the range of the given function (so it looks like something is left for you to do after all). This job is quite straightforward since you know that f is a one-to-one function ....