Prove that if g(x) is even then the composition f(g(x) is also even, whether or not f is even or odd
Let g(x) be an even function, meaning g(-x) = g(x) for all x.
Now, let's consider the composition f(g(x)). Since g(x) is even, we know that g(-x) = g(x) for all x.
Therefore, when we substitute -x into g(x), we get the same output as when we substitute x into g(x). Thus, f(g(-x)) = f(g(x)).
This means that f(g(x)) is also an even function, as f(-x) = f(x) for all x.
Therefore, if g(x) is even, then the composition f(g(x)) is also even, regardless of whether f is even or odd.