Answer
(a) See the explanation
(b) See the explanation
(c) See the explanation
(d) $f(x)=\underbrace{\frac{1}{2}[(2^x+(x-3)^2)+(2^{-x}+(-x-3)^2)]}_{even}+\underbrace{\frac{1}{2}[(2^x+(x-3)^2)-(2^{-x}+(-x-3)^2)]}_{odd}$
Work Step by Step
(a) Recall: $E(x)$ is an even function if $E(-x)=E(x)$.
Show that $E(x)$ is an even function:
$E(-x)=f(-x)+f(-(-x))=f(-x)+f(x)=f(x)+f(-x)=E(x)$
Thus, $E(x)$ is an even function.
(b) Recall: $O(x)$ is an odd function if $O(-x)=-O(x)$.
Show that $O(x)$ is an odd function:
$O(-x)=f(-x)-f(-(-x))=f(-x)-f(x)=-(f(x)-f(-x))=-O(x)$
Thus, $O(x)$ is an odd function.
(c) Let $f(x)$ be a function. From part (a) and (b), we have shown that $E(x)=f(x)+f(-x)$ is an even function and $O(x)=f(x)-f(-x)$ is an odd function. Consequently, $(\frac{1}{2}E)(x)$ is an even function and $(\frac{1}{2}O)(x)$ is an odd function.
Notice that
$f(x)=\frac{1}{2}\cdot 2f(x)$
$=\frac{1}{2}\cdot [(f(x)+f(-x))+(f(x)-f(-x))]$
$=\frac{1}{2}\cdot (f(x)+f(-x))+\frac{1}{2}\cdot (f(x)-f(-x))$
$=\frac{1}{2}\cdot E(x)+\frac{1}{2}\cdot O(x)$
$=(\frac{1}{2}E)(x)+(\frac{1}{2}O)(x)$
Thus, $f(x)$ can be written as the sum of an even function $(\frac{1}{2}E)(x)$ and an odd function $(\frac{1}{2}O)(x)$.
(d) Using the result in part (c), take $E(x)=f(x)+f(-x)=(2^x+(x-3)^2)+(2^{-x}+(-x-3)^2)$ and $O(x)=f(x)+f(-x)=(2^x+(x-3)^2)-(2^{-x}+(-x-3)^2)$.
Then, $f(x)$ can be written as the sum of an even function $\frac{1}{2}[(2^x+(x-3)^2)+(2^{-x}+(-x-3)^2)]$ and an odd function $\frac{1}{2}[(2^x+(x-3)^2)-(2^{-x}+(-x-3)^2)]$.
