Exercise 1.3.70

Question

Suppose $g$ is an odd function and let $h=f \circ g$. Is $h$ always an
odd function? What if $f$ is odd? What if $f$ is even?

Khagan Gulusoy · Accepted Answer

\begin{itemize}\item ($f$ is an odd function) Let $X$ be the domain of $f \circ g$. We have: $$\forall x \in x: f \circ g(x)=f(g(x)).$$ Since $g$ is an odd function, we have:
$$
\forall x \in X: f(g(x))=f(-g(-x)).
$$
Since $f$ and $g$ are odd functions, we have:
$$\forall x \in X: f(-g(-x))=-f(g(-x))=-f(-g(x)).$$
Therefore, we have.
$$
\forall x \in X: f(g(x))=-f(-g(x)).
$$
Thus, in this case, $f \circ g$ is an odd function.

\item ($f$ is an even function) Let $X$ be the domain of $f \circ g$. We have:
$$
\forall x \in X: f \circ g(x)=f(g(x))=f(-g(-x))=f(g(-x))=f(-g(x)) 	ext {. }
$$
Therefore, we have:
$$
\forall x \in X: f(g(x))=f(-g(x)) 	ext {, }
$$
which means that in this case $f \circ g$ is even.
\end{itemize}

Exercise 1.3.70

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Exercise 1.3.70

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