Chapter 2.5, Problem 91E

To determine

To prove : that if $f$ and $g$ are both even and odd function, then the sum if one is odd and one is even

Explanation of Solution

Given information : The question is

  $f+g$ Necessarily even and if both are odd, is their sum necessarily odd

Proof : $f\left(x\right)$ and $g\left(x\right)$ are even function

  $\because f\left(-x\right)=f\left(x\right)$ and $g\left(-x\right)=g\left(x\right)$

  $g\left(-x\right)+f\left(-x\right)=g\left(x\right)+f\left(x\right)$

So, the sum of two even function is necessary even function

  $f\left(x\right)$ And $g\left(x\right)$ are odd function

  $\begin{array}{l}\because f\left(-x\right)=-f\left(x\right)\&g\left(-x\right)=-g\left(x\right)\\ g\left(-x\right)+f\left(-x\right)=-g\left(x\right)+\left(-f\left(x\right)\right)\end{array}$

So, The sum of two odd function is necessary odd function

Let $f\left(x\right)$ is even function

And $g\left(x\right)$ is odd function

  $\therefore g\left(-x\right)+f\left(-x\right)=-g\left(x\right)+f\left(x\right)$

So, the sum of even and odd function is neither even or nor odd