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

• MHB
• karush
In summary, the conversation discusses finding the value of $g'(1)$, the derivative of the inverse function $g$ of the function $f(x)=(2x+1)^3$, given that $f(0)=1$. The conversation uses the fact that if $f(x)=y$, then $g'(y)=\dfrac{1}{f'(x)}$ to calculate $g'(1)$, which is equal to $\dfrac{1}{6}$ or option (D) on the provided multiple choice options.
karush
Gold Member
MHB
$\tiny{219\quad inverse function}$
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

Last edited:
karush said:
$\tiny{219\quad inverse function}$
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 $g'(1) = \dfrac16$.

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

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

sorry... well more into tho

## What is the meaning of "219 GRE"?

"219 GRE" refers to a specific standardized test called the Graduate Record Examination (GRE). It is used as an admissions requirement for many graduate schools in the United States.

## What does g'(1) mean?

g'(1) is a mathematical notation that represents the derivative of the function g(x) at the point x=1. In other words, it is the rate of change of the function at the specific point x=1.

## Why is the value of g'(1) important?

The value of g'(1) can provide important information about the behavior of the function g(x) at the point x=1. It can tell us whether the function is increasing or decreasing at that point, and the steepness of the curve at that point.

## How is g'(1) calculated?

g'(1) is calculated using the rules of differentiation, which involve finding the slope of the tangent line to the curve at the point x=1. This can be done using techniques such as the power rule, product rule, and chain rule.

## What factors can affect the value of g'(1)?

The value of g'(1) can be affected by the shape and behavior of the function g(x) near the point x=1. It can also be influenced by the choice of mathematical techniques used to calculate the derivative, as well as any errors or approximations made during the calculation.

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