Posted by xavier on .
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,
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