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

  1. Sep 10, 2013 #1
    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?
     
  2. jcsd
  3. Sep 10, 2013 #2
    General case: [itex]f[/itex] is even, i.e. [itex]f(x)=f(-x)[/itex] for every [itex]x[/itex]. Thus [itex]g(f(x))=g(f(-x))[/itex], as both involve putting the the same input into [itex]g[/itex].

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