- #1

karush

Gold Member

MHB

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$\tiny{219\quad inverse function}$

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

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

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