Ok so
if f is even and g is even then fog is?
f(x) = f(-x)
g(x) = g(-x)
f(g(x)) = f(g(-x))
f(-g(x)) = f(g(x))
f(-g(x)) = f(-g(-x))
so f(-g(x)) = f(g(x) fog is even???
Hello, Fabio010!
This is simpler than you think . . .
$\displaystyle \text{If }f\text{ is even and }g\text{ is even, then }f\!\circ\!g\text{ is?}$
We know that: .$\displaystyle \begin{Bmatrix}f(\text{-}x) &-& f(x) \\ g(\text{-}x) &=& g(x)\end{Bmatrix}$
The question is: .$\displaystyle \begin{Bmatrix}\text{Is }f\!\circ\!g \text{ even?} \\ \text{Does }f(g(\text{-}x)) \text{ equal }f(g(x))\,? \end{Bmatrix}$
Since $\displaystyle g(\text{-}x) = g(x)$, we have: .$\displaystyle f(g(\text{-}x)) \:=\:f(g(x))$ . . . . Yes!
If $\displaystyle f$ is odd then $\displaystyle f(-x)=-f(x)$, if $\displaystyle g$ is odd then $\displaystyle g(-x)=-g(x)$.
Show: $\displaystyle f(g(-x))=-f(g(x))$
$\displaystyle f(g(-x))=f(-g(x))=-f(g(x))$
Therefore $\displaystyle f\circ g$ is odd.