I know it is a simple problem but I am having trouble. Here is what I have so far:
Let $f(x) = x^5 + 2x^3 + x - 1$
a) Find $f(1)$ and $f'(1)$
I have a) done. $f(1)$ is $3$ and $f'(1)$ is $12$
b) Find $f^{-1}(3)$ and $(f^{-1})'(3)$
I need help with the first part. I think the way to find the inverse is to switch the $x$'s with $y$'s and then solve for $y$. But I am having trouble completing this. I have the following: $$ x = y^5 + 2y^3 + y -1 $$ $$ x - 1 = y^5 + 2y^3 + y $$ What am I supposed to do here?
I don't think you need to find $f^{-1}$ explicitly. Remember that $f^{-1}(3)$ in this case is the number $x$ such that $f(x)=3$, and use what you've found in part (a).
For $(f^{-1})' (3)$, remember that by the inverse function theorem, $$ (f^{-1})'(b)=\frac{1}{f'(a)} $$ where $f(a)=b$, and again use what you've found in part (a).