True. Because f(1) = f(-1) so it is not one-to-one.

False, consider f(x) = x.24. If a function is odd, then it does not have an inverse function.

f(g(x)) = f(g(-x))28. If g(x) is an even function then f(g(x)) is even for every

function f(x).

Is that true?

Let f(x) = x^2 and g(x) = x+1.29. If f(x) is an even function then f(g(x)) is even for every

function g(x).

Then is,

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