For the following:
If $f$ is an odd function, then $|f|$ is _______.
If $f$ is an odd function and $g$ is an even function then, $(f\circ g)(x)$ is _______.
If $f$ is an odd function and $g$ is an odd function, then $(f\cdot g)(x)$ is _______.
I can't tell the answer all of a sudden, however whenever I start trying to prove them, I get it.
Anyways, how would you guys approach this? I learned how to actually go about proving it but is there anyway you can solve this without actually trying to prove it?
Best Answer
1. If $f$ is odd , then $|f|$ is even because $\left | f(-x) \right |= \left | -f(x) \right |= \left | f(x) \right |$.
2. If $f$ is odd and $g$ is even then $f\circ g$ is even. That is because if $h(x)=(f\circ g)(x)$ then:
$$h(-x)=\left ( f\circ g \right )(-x)= f\left ( g(-x) \right )= f(g(x))=h(x)$$
3. If $f$ is odd and $g$ is odd then $\varphi(x)=f(x) \cdot g(x)$ is even. That is because:
$$\varphi(-x)= f(-x) g(-x)= -f(x)\cdot (-g(x))= f(x)g(x)=\varphi(x)$$
Fixed typos!