219 AP Calculus Exam Inverse function

[karush]

Let $f(x)=(2x+1)^3$ and let g be the inverse of $f$. Given that $f(0)=1$, what is the value of $g'(1)?$


$(A)\, \dfrac{2}{27} \quad
(B)\, \dfrac{1}{54} \quad
(C)\, \dfrac{1}{27} \quad
(D)\, \dfrac{1}{6} \quad
(E)\, 6$


ok not sure what the best steps on this would be but assume we first find $f^{-1}(x)$


so rewrite at
$y=(2x+1)^3$
exchange x for y and y for x
$x=(2y+1)^3$
Cube root each side
$\sqrt[3]{x}=2y+1$
isolate y
$\dfrac{\sqrt[3]{x}-1}{2}=y=f'(x)$


so then...

 

[skeeter]

let $g(x)$ be the inverse of $f(x)$

$f(0) = 1 \implies g(1) = 0$

property of an inverse function ...

$f[g(x)] = x$

derivative of the above equation ...

$f'[g(x)] \cdot g'(x) = 1 \implies g'(x) =\dfrac{1}{f'[g(x)]} \implies g'(1) = \dfrac{1}{f'[g(1)]} = \dfrac{1}{f'(0)}$

$f'(x) = 6(2x+1)^2 \implies f'(0) = 6 \implies g'(1) = \dfrac{1}{6}$