
William L. answered 10/23/22
Mathematics PhD from Cambridge with 20 years of experience!
Looking at h(x) = (x + 2)8 , and f(x) = x8 , we can see that g(x) = x + 2 works, since
f o g (x) = f(g(x)) = f(x + 2) = (x + 2)8.
If it's not clear how to get to g(x) = x + 2, just note that f has a partial inverse f-1(y) = y1/8 , which is the inverse of f(x) when we restrict the variable x to x ≥ 0. In this case, when x + 2 ≥ 0, we have:
f-1(f(g(x))) = g(x) , since the f and the f-1 cancel out. We also have:
f-1(f(g(x))) = (f(g(x)))1/8 = ((x+2)8)1/8 = (x + 2) , since x + 2 ≥ 0. Note that, in general (y8)1/8 = |y|.
The reasoning above only works for x+2 ≥ 0, since only in this case can we use f-1 as we had defined, but we can set g(x) = x + 2 and check that it works even if x + 2 < 0.