- #1

karush

Gold Member

MHB

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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...

(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...

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