Inverse equations

**Arez** · Apr 26th 2009, 12:14 PM

it gives this:

The function $f(x) = e^{x^5 + x}$ is strictly increasing on $(-infinity, infinity)$ and therefore has an inverse function $y = f^{-1}(x)$ .

(a) Find $f^{-1}(e^2)$

(b) Find $(f^{-1})'(e^2)$

i know the rule is $(f^{-1})'(y) = 1/f'(x)$ for the second, but what do i plug into $f(x) = e^{x^5 + x}$ to make it $e^2$ ??

**Jhevon** · Apr 26th 2009, 12:19 PM

Originally Posted by Arez

... but what do i plug into $f(x) = e^{x^5 + x}$ to make it $e^2$ ??

er, x = 1 seems to work

it should be more or less easy to see this, but if not, just set x^5 + x = 2 and solve for x. which is harder.

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