Question

Given the function g(x) = 41x3 + a for some constant a, which describes the inverse function g–1(x)?

no restriction on the domain of g(x); inverse is g–1(x) = RootIndex 3 StartRoot StartFraction x minus a Over 41 EndFraction EndRoot
no restriction on the domain of g(x); inverse is g–1(x) = (StartFraction x minus a Over 41 EndFraction) cubed
domain of g(x) restricted to x ≥ 0; inverse is g–1(x) = RootIndex 3 StartRoot StartFraction x minus a Over 41 EndFraction EndRoot
domain of g(x) restricted to x ≥ 0; inverse is g–1(x) = (StartFraction x minus a Over 41 EndFraction) cubed

Answer 1

It's A

Step-by-step explanation:

Answer 2

A. No restriction on the domain of g(x); inverse is g–1(x) = RootIndex 3 StartRoot StartFraction x minus a Over 41 EndFraction EndRoot

Step-by-step explanation:

Given the function
`g(x)=41x^3+a\\`, we are to find the inverse of the function a shown below:

Ley y = g(x)

`y=41x^3+a\\`

replace y with x

`x=41y^3+a\\`

make y the subject of the formula:

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

Replace y with
`g^(-1)(x)`

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

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

No restriction on the domain of g(x). Hence the correct option is A