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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?

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  • $\begingroup$ $[f^{-1}]'(a) = \frac{1}{f'(f^{-1}(a))}$. So you just need to figure out what $f^{-1}(2)$ is and evaluate $f'$ at that point. Does this help? $\endgroup$ Oct 20, 2017 at 13:48
  • $\begingroup$ my teacher gave hint to use Roll's theorem. $\endgroup$
    – Tortoise
    Oct 20, 2017 at 14:15
  • $\begingroup$ Hmm...I'm not sure how one would use Rolle's theorem to solve this. Sorry. $\endgroup$ Oct 20, 2017 at 14:57

1 Answer 1

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\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}

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