# -apc.1.1.5 inverse function

• MHB
• karush
In summary, the inverse of $f(x) = (2x+1)^3$ is $g(x) = \dfrac{\sqrt[3]{x}-1}{2}$ and the value of $g'(1)$ is $\dfrac{1}{6}$.

#### karush

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

ok not real sure what the answer is but I did this (could be easier I am sure}

rewrite as
$y=(2x+1)^3$
exchange x and rename y to g
$x=(2g+1)^3$
Cube root each side
$\sqrt[3]{x}=2g+1$
isolate g
$g=\dfrac{\sqrt[3]{x}-1}{2}$
so
$\left(\dfrac{\sqrt[3]{x}-1}{2}\right)' =\dfrac{1}{6x^{\dfrac{2}{3}}}$
apply $x=1$
$\dfrac{1}{6(1)^{\dfrac{2}{3}}}=\dfrac{1}{6}$

hopefully

$f(x) = (2x+1)^3 \implies f'(x) = 6(2x+1)^2$

$f$ and $g$ are inverses tells us two things ...

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

(2) $f[g(x)] = x$

take the derivative of equation (2) ...

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

$g'(1) = \dfrac{1}{f'[g(1)]} = \dfrac{1}{f'(0)} = \dfrac{1}{6}$

skeeter said:
$f(x) = (2x+1)^3 \implies f'(x) = 6(2x+1)^2$

$f$ and $g$ are inverses tells us two things ...

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

(2) $f[g(x)] = x$

take the derivative of equation (2) ...

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

$g'(1) = \dfrac{1}{f'[g(1)]} = \dfrac{1}{f'(0)} = \dfrac{1}{6}$
Mahalo

btw how you rate this problem Easy, Medium, or Hard

Medium

noted thanks

skeeter said:
$f(x) = (2x+1)^3 \implies f'(x) = 6(2x+1)^2$

$f$ and $g$ are inverses tells us two things ...

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

(2) $f[g(x)] = x$

take the derivative of equation (2) ...

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

$g'(1) = \dfrac{1}{f'[g(1)]} = \dfrac{1}{f'(0)} = \dfrac{1}{6}$

that helped a lot