## Calculus 8th Edition

When $f$ is odd, we can find that: $f(g(-x)) = f(-g(x))= -f(g(x)) = -h(x)$ Therefore $h$ is odd. However, when $f$ is even, we can find that: $f(g(-x)) = f(-g(x))= f(g(x)) = h(x)$ Therefore $h$ is even. Thus, $h$ is not always an odd function.