Posted by **xavier** on Wednesday, October 3, 2012 at 11:43pm.

Auppose f is even but g is an arbitrary function (possibly neither even nor odd).

(a) Is f o g still even? Show why or why not.

(b) Is g o f even? Show why or why not.

- Math -
**Steve**, Thursday, October 4, 2012 at 10:35am
fog is even, since f is even.

since f(-x) = f(x), f(-g) = f(g)

gof is unknown. for example,

f(x) = x^2

g(x) = x

g(f) = f = x^2 is even

g(x) = √x

g(f) = √f = x is odd

g(x) = √x + 1

g(f) = √f + 1 = x+1 neither even nor odd

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