Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $ Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $
How do I go about solving a problem like this? What are the steps? 

first you need to find $x$  such that $f(x) = 4.$  you know there is exactly one such $x$ because $f$ has an inverse. by trial and error, you find that $x = 0.$  so $f$  sends $0$  to $4.$  the slope of $f$  at $x = 0$ is $f'(0) = 7.$  therefore the slope of $f^{-1}$ at the point $y = 4$ is $\left(f^{-1}\right)'(4) = \frac17,$  the reciprocal of the slope of $f$ at the point $x = 0.$
