Prove or disprove: for every $f,g : \mathbb R\to\mathbb R$ even, the composition $h= f\circ g$ is even.

Question

Proof:

Given $f,g:\mathbb R\to\mathbb R$ even
then $f(-x)=f(x)$ and $g(-x)=g(x)$

then $h=f \circ g$
then $h=f(g(x))=f(g(-x))$
then $h=g(-x)=g(x)$
since $x \neq -x$ the composition of two even functions is even?

Im having trouble writing out these types of proofs and how to go about them. Does my proof make sense? Thanks.

Answer 1

$f(g(-x)) = f(g(x))$ since $g$ is even, therefore $f\circ g$ is even. Note: we only needed that $g$ was even.