# Why will a composition function of an even function always be even?

#### robertmatthew

I apologize in advance if the answer to this is really simple; I often overlook simple solutions when something trips me up.

For example, if f(x)=x2 and g(x)=x3/2, and g(f(x)) is therefore, after simplification, x3, why is that still an even function if x3 graphed under other circumstances is odd?

#### economicsnerd

General case: $f$ is even, i.e. $f(x)=f(-x)$ for every $x$. Thus $g(f(x))=g(f(-x))$, as both involve putting the the same input into $g$.

Specific example: You made a computational error. You have functions $f:\mathbb R \to \mathbb R_+$ and $g:\mathbb R_+ \to \mathbb R_+$, so indeed:
- The composition $g\circ f:\mathbb R \to \mathbb R_+$ makes sense.
- $f$, and thus $g\circ f$ as well, has domain $\mathbb R$. So it makes sense to ask whether $f$ and $g\circ f$ are even.
You can check that in fact $g(f(x))= \bigg(f(x)\bigg)^{\frac32} = \bigg(\sqrt{f(x)}\bigg)^3= \bigg(\sqrt{x^2}\bigg)^3 = |x|^3$.

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