Question

Precalculus question, image attached.

Answer 1

g(x)= x+1

Explanation:

`h(x)= (fog)(x)= f(g(x))`

Given
`f(x)= \sqrt[3]{x+2}`,
`f(x)= \sqrt[3]{x+3}`

`f(g(x))= \sqrt[3]{x+3}`

`f(x)= \sqrt[3]{x+2}`

Replace x with g(x)

`f(g(x))= \sqrt[3]{g(x)+2}`

Now we need to find out what we plug in for g(X) to get x+3 under the radical

g(x) +2 is under the radical. we need x+3 , so we replace g(x) with x+`

So g(x)+2 becomes x+1+2 is x+3

Hence g(x) is x+1