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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