Ok so

if f is even and g is even then fog is?

f(x) = f(-x)

g(x) = g(-x)

f(g(x)) = f(g(-x))

f(-g(x)) = f(g(x))

f(-g(x)) = f(-g(-x))

so f(-g(x)) = f(g(x) fog is even???

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- Nov 26th 2011, 04:09 PMFabio010Even and odd composite functions.
Ok so

if f is even and g is even then fog is?

f(x) = f(-x)

g(x) = g(-x)

f(g(x)) = f(g(-x))

f(-g(x)) = f(g(x))

f(-g(x)) = f(-g(-x))

so f(-g(x)) = f(g(x) fog is even??? - Nov 26th 2011, 04:57 PMPlatoRe: Even and odd composite functions.
- Nov 26th 2011, 05:12 PMSorobanRe: Even and odd composite functions.
Hello, Fabio010!

This is simpler than you think . . .

Quote:

$\displaystyle \text{If }f\text{ is even and }g\text{ is even, then }f\!\circ\!g\text{ is?}$

We know that: .$\displaystyle \begin{Bmatrix}f(\text{-}x) &-& f(x) \\ g(\text{-}x) &=& g(x)\end{Bmatrix}$

The question is: .$\displaystyle \begin{Bmatrix}\text{Is }f\!\circ\!g \text{ even?} \\ \text{Does }f(g(\text{-}x)) \text{ equal }f(g(x))\,? \end{Bmatrix}$

Since $\displaystyle g(\text{-}x) = g(x)$, we have: .$\displaystyle f(g(\text{-}x)) \:=\:f(g(x))$ . . . . Yes!

- Nov 27th 2011, 04:30 AMFabio010Re: Even and odd composite functions.
So if f is odd and g is odd then:

fog is odd because:

f(g(x)) not equal f(-g(x))

since g(-x) = -g(x)

and -f(g(x)) = f(-g(x))

right? - Nov 27th 2011, 04:39 AMSironRe: Even and odd composite functions.
If $\displaystyle f$ is odd then $\displaystyle f(-x)=-f(x)$, if $\displaystyle g$ is odd then $\displaystyle g(-x)=-g(x)$.

Show: $\displaystyle f(g(-x))=-f(g(x))$

$\displaystyle f(g(-x))=f(-g(x))=-f(g(x))$

Therefore $\displaystyle f\circ g$ is odd. - Nov 27th 2011, 05:38 AMFabio010Re: Even and odd composite functions.
Ok thanks for the help.