Question

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

Answer 1

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

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,\\(x-a)/(41)=y^3,\\y=\sqrt[3]{(x-a)/(41)}`.

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

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