Thread: Inverse equations

it gives this:

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

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

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

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

Originally Posted by Arez

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

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.