Question

Given the function g(x)=41x^3+a for some constant a, which describes the inverse function g^-1(x)

Answer 1

Answer:

$g^{-1}(x)=\sqrt[3]{\frac{x-a}{41}}$

Step-by-step explanation:

The inverse of a function has $x$ and $y$ values switched from the original function. Therefore, simply switch $x$ and $y$ and isolate $y$ to get your inverse function:

Original function: $g(x)=41x^3+a$

Switching $x$ and $y$, then isolating $y$:

$x=41y^3+a,\\x-a=41y^3,\\\frac{x-a}{41}=y^3,\\y=\sqrt[3]{\frac{x-a}{41}}$.

Therefore, the inverse of the $g(x)=41x^3+a$ is:

$\fbox{ g^{-1}(x)=\sqrt[3]{\frac{x-a}{41}} }$.