For the benefit of other users, I've transcribed your image:
Find the inverse functions of each one-to-one function:
A. \(\displaystyle f(x) = (x-3)^3\)
B. \(\displaystyle g(x) = x^2 - 2x + 5\), where \(\displaystyle x \ge 1\)
C. \(\displaystyle h(x) = 2cos(2x) - 3\), where \(\displaystyle 0 \le x \le \pi\)
D. \(\displaystyle f(x) = 3^{x-3} + 2\)
With that out of the way... in the future, if you're ever unsure of an answer you can always check it yourself. In this case, we'll use the definition of inverse functions, that \(\displaystyle f(f^{-1}(x)) = f^{-1}(f(x)) = x\). I'll work through checking one answer and leave you to check the rest.
\(\displaystyle f(f^{-1}(x)) = (\left[ \sqrt[3]{x} + 3 \right]-3)^3\)
\(\displaystyle = \left[ \sqrt[3]{x} + 3 \right]^3 - 9\left[ \sqrt[3]{x} + 3 \right]^2 + 27\left[ \sqrt[3]{x} + 3 \right] - 27\)
\(\displaystyle = \left[ 9x^{2/3} + 27x^{1/3} + x + 27 \right] - \left[ 9x^{2/3} + 54x^{1/3} + 81 \right] + \left[ 27x^{1/3} + 81 \right] - 27\)
\(\displaystyle = 9x^{2/3} + 27x^{1/3} + x + 27 - 9x^{2/3} - 54x^{1/3} - 81 + 27x^{1/3} + 81 - 27\)
\(\displaystyle = x\)
Okay, that's good. Let's check the other way around just to be sure though:
\(\displaystyle f^{-1}(f(x)) = \sqrt[3]{(x-3)^3} + 3\)
\(\displaystyle = (x-3) + 3\)
\(\displaystyle = x\)
This also checks out, so therefore we know you're answer is correct.