- #1

- 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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter robertmatthew
- Start date

- #1

- 48

- 0

For example, if f(x)=x

- #2

- 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].

Share: