219 GRE what is the value of g'(1)

[karush]

$\tiny{219\quad inverse function}$
Let $f(x)=(2x+1)^3$ and let g be the inverse of $f$. Given that $f(0)=1$, what is the value of $g'(1)?$
$(A)\, \dfrac{2}{27} \quad
(B)\, \dfrac{1}{54} \quad
(C)\, \dfrac{1}{27} \quad
(D)\, \dfrac{1}{6} \quad
(E)\, 6$
so rewrite at
$\quad y=(2x+1)^3$
exchange x and y
$\quad x=(2y+1)^3$
Cube root each side
$\quad \sqrt[3]{x}=2y+1$
isolate y to get g(x)
$\quad \dfrac{\sqrt[3]{x}-1}{2}=y=g(x)$
then
$\quad \left(\dfrac{x^{1/3}-1}{2}\right)'
=\dfrac{1}{6 x^{2/3}}=g'(x)$
so then
$\quad g'(1)=\dfrac{1}{6 (1)^{2/3}}=\dfrac{1}{6}\quad (D) $ok for a GRE test question I thot this was more caculation than neededI don't know the given answer but presume this is correct

 

[Opalg]

$\tiny{219\quad inverse function}$
Let $f(x)=(2x+1)^3$ and let g be the inverse of $f$. Given that $f(0)=1$, what is the value of $g'(1)?$
$(A)\, \dfrac{2}{27} \quad
(B)\, \dfrac{1}{54} \quad
(C)\, \dfrac{1}{27} \quad
(D)\, \dfrac{1}{6} \quad
(E)\, 6$
so rewrite at
$\quad y=(2x+1)^3$
exchange x and y
$\quad x=(2y+1)^3$
Cube root each side
$\quad \sqrt[3]{x}=2y+1$
isolate y to get g(x)
$\quad \dfrac{\sqrt[3]{x}-1}{2}=y=g(x)$
then
$\quad \left(\dfrac{x^{1/3}-1}{2}\right)'
=\dfrac{1}{6 x^{2/3}}=g'(x)$
so then
$\quad g'(1)=\dfrac{1}{6 (1)^{2/3}}=\dfrac{1}{6}\quad (D) $ok for a GRE test question I thot this was more caculation than neededI don't know the given answer but presume this is correct

You could do it a bit more easily by using the fact that if $f(x)=y$ then $g'(y) = \dfrac1{f'(x)}$.

Since $f(0) = 1$ and $f'(0) = 6(2x+1)^2\Big|_{x=0} = 6$, it follows that $g'(1) = \dfrac16$.

 

[skeeter]

you've posted this exact same problem once before ...

https://mathhelpboards.com/calculus-10/219-ap-calculus-exam-inverse-function-26440.html

 

[karush]

sorry... well more into tho