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

Answer:

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

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:

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

Replace y with $g^{-1}(x)$

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

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

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

Answer 2

Answer:

the answer is A

Explanation:

i did the test already on edg