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)=x^3/2, 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.