I am super stuck on a problem. I am on the chapter of my calculus book on the derivatives on inverse functions but I can't figure it out.
[Note: I am writing the inverse of $f$ as $f^{-1}$] The problem is:
Find $(f^{-1})'(5)$ where $f(x) = 3x^3 + 4x^2 + 6x + 5$.
The back of the book says the answer is $1/6 $
I know that $(f^{-1})'(a)$ = $1 / f'( f^1(a) )$ I used symlab to find $f(x) = 5$ and it gives me: $-(2/3) * (+ or -) i(\sqrt{14} / 3)$ which is a complex number so that can't be what I'm looking for.
Then i tried implicit derivation to find $dy/dx$ and that just gave me: $9x^2 + 8x + 6$ And plugging in $5$ gives a number way too big.
I have absolutely no idea what I'm doing wrong. I've watched all kinds of videos on this and reread the chapter a bunch of times and still can't get it. They always find the equation (f^-1). But if you just find that equation, then why not just take it's derivative rather than the weird [ $1 / f'( f^{-1}(x))$ ] ? I must be missing some basic idea of this, but I don't know what.
I really appreciate any help! Thanks!
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^{-1}:\mathbb{R}\to \mathbb{R}$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^{-1}$ at $5$?