- #1

robertmatthew

- 48

- 0

For example, if f(x)=x

^{2}and g(x)=x

^{3/2}, and g(f(x)) is therefore, after simplification, x

^{3}, why is that still an even function if x

^{3}graphed under other circumstances is odd?

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- Thread starter robertmatthew
- Start date

- #1

robertmatthew

- 48

- 0

For example, if f(x)=x

- #2

economicsnerd

- 269

- 24

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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