Finding inverse of a real valued function

Question

Let $f(x)$ be a real-valued function defined on the interval $(-5, 5)$ such that

$$e^{-x}f(x) = 2 + \int\limits_0^x (t^4 + 1)^{1/2} \ dt$$

for all $x \in (-5, 5)$. Let $f^{-1}(x)$ be the inverse function of $f(x)$. Find $(f^{-1})^{'}(2)$

This question is part of a calculus course.

Please give me some hints on how to approach this question?

Answer 1

\begin{align} f(x) &= 2\exp(x) + \exp(x)\int\limits_0^x \sqrt{t^4 + 1} \ dt ,\\ f'(x)&=\exp(x) \left( 2+\int _{0}^{x}\sqrt{t^4+1}\, dt \right) +\exp(x)\sqrt{x^4+1} . \end{align}

Note that at $x=0$ that nasty integral vanishes and $f(0)=2$ hence $f^{-1}(2)=0$.

This is all you need together with the hint you already got,

\begin{align} \left[f^{{-1}}\right]'(a)={\frac {1}{f'\left(f^{{-1}}(a)\right)}} . \end{align}