Odd and even properties of functions

Question

Could I get a hint or a clue on how to solve this problem?

Problem:

($f$ is even if $f(-x)=f(x)$ and $f$ is odd if $f(-x)=-f(x)$.)

Suppose $f$ is a function defined on all of $\mathbb{R}$

(a) Check whether $g(x) = \frac{(f(x))+(f(-x))}{2}$ is even, odd or neither.

(b) Use (a) as inspiration to show that you can always write a function $f$ as a sum of an odd and an even function

So since $f$ is a function defined on all of $\mathbb{R}$, that would imply that $f$ is not even? Right? Like if $f(x)=x^2$ (an even function) then $f$ would not be defined on on all of $\mathbb{R}$. So should I assume that $f$ is odd or neither? And where do I go on from there??

Answer 1

First, for checking whether or not $g(x)$ is even or odd, plug in $-x$ for $x$, or equivalently, $x\mapsto -x$. The function is even because $g(x)=g(-x)$.

Second, $$\begin{align*}f(x)&=\frac{f(x)}{2}+\frac{f(x)}{2}\\&=\frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}\\&=\frac{f(x)}{2}+\frac{f(-x)}{2}+\frac{f(x)}{2}-\frac{f(-x)}{2} \end{align*}$$

Can you finish from here?

Answer 2

Note that a) is clear and for b) you can see that if $g$ is a function defined on all of $\mathbb{R}$ and $f$ is defined on all $\mathbb{R}$ we can see that $$\boxed{g(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{even function}+\underbrace{\frac{f(x)-f(-x)}{2}}_{odd function}.}$$