Question

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.

Answer 1

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