Ok so if f is even and g is even then fog is? f(x) = f(-x) g(x) = g(-x) f(g(x)) = f(g(-x)) f(-g(x)) = f(g(x)) f(-g(x)) = f(-g(-x)) so f(-g(x)) = f(g(x) fog is even???
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Originally Posted by fabio010 ok so if f is even and g is even then fog is? F(x) = f(-x) & g(x) = g(-x) f(g(x)) = f(g(-x)) f(-g(x)) = f(g(x)) f(-g(x)) = f(-g(-x)) so f(-g(x)) = f(g(x) fog is even??? correct.
Hello, Fabio010! This is simpler than you think . . . We know that: . The question is: . Since , we have: . . . . . Yes!
So if f is odd and g is odd then: fog is odd because: f(g(x)) not equal f(-g(x)) since g(-x) = -g(x) and -f(g(x)) = f(-g(x)) right?
If is odd then , if is odd then . Show: Therefore is odd.
Ok thanks for the help.