Answer :
f(x) = x³ + 4 and g(x) = ∛(x - 4)
so, the first thing the question asks of you is to see that f(g(x)) = x. try it out, plugging g(x) in as the x-variable in the first equation:
f(g(x)) = (∛(x - 4))³ + 4
they were merciful in writing this problem, and thankfully your cube roots cancel out and don't cause you any trouble. continue solving:
f(g(x)) = (∛(x - 4))³ + 4 ... cube root and exponent cancel
f(g(x)) = x - 4 + 4 ... simplify
f(g(x)) = x
so, yep. that one worked. try out the second half of the question: g(f(x)) = x
g(f(x)) = ∛((x³ + 4) - 4) ... simplify inside your radical, cancelling out the 4s
g(f(x)) = ∛(x³) ... the cube root of x³ is x itself, so:
g(f(x)) = x
and there you are. you've confirmed that these are inverses.
so, the first thing the question asks of you is to see that f(g(x)) = x. try it out, plugging g(x) in as the x-variable in the first equation:
f(g(x)) = (∛(x - 4))³ + 4
they were merciful in writing this problem, and thankfully your cube roots cancel out and don't cause you any trouble. continue solving:
f(g(x)) = (∛(x - 4))³ + 4 ... cube root and exponent cancel
f(g(x)) = x - 4 + 4 ... simplify
f(g(x)) = x
so, yep. that one worked. try out the second half of the question: g(f(x)) = x
g(f(x)) = ∛((x³ + 4) - 4) ... simplify inside your radical, cancelling out the 4s
g(f(x)) = ∛(x³) ... the cube root of x³ is x itself, so:
g(f(x)) = x
and there you are. you've confirmed that these are inverses.