Given the functions below what value does a need to be for them to be inverses? 1/x x^-1

Question

h(x)=√x^3-2 g(x)=x^3+a

Answer 1

Sarah,

First, we can express the inverse of g(x)

y = g(x) = x^3 + a

by swapping the letters x and y

x = y^3 + a

then solving for y

x - a = y^3

y = (x - a)^(1/3) is the inverse of g(x)

But the problem statement seems impossible, as it’s hard to imagine that the cube of a square root could be the inverse of a cube

h(x)=√x^3-2

I wonder whether the original problem was different from what you’ve asked about:

h(x) = ³√x - 2 or even ³√(x-2)

this latter expression could be written as

(x-2)^(1/3), which is exactly the form of g inverse created above.

so, assuming h is a cube root, then setting a to 2 is all that’s necessary.