Chapter 6.4, Problem 51HP

To determine

To explain: even degree polynomial function is also an even function and every domain of $\text{x}$ for $\text{f}\left(\text{-x}\right)\text{=f}\left(\text{x}\right)$ .

Answer to Problem 51HP

Every even degree polynomial not even function.

Explanation of Solution

Given:

  $\text{f}\left(\text{-x}\right)\text{=f}\left(\text{x}\right)$ .

Concept used:

For even function the equation of the function will be:

  $\text{n=2k}$

  $\text{f}\left(\text{x}\right){\text{=a}}_{\text{2k}}{\text{x}}^{\text{2k}}{\text{+a}}_{\text{2k-1}}{\text{x}}^{\text{2k-1}}\text{+}\mathrm{......}{\text{+a}}_{\text{1}}{\text{x+a}}_{\text{0}}$ .

For a function to be even only if:

  $\text{f}\left(\text{-x}\right)\text{=f}\left(\text{x}\right)$ .

Calculation:

First study an even -degree polynomial function is even function.

  $\text{n=2k}$

  $\text{f}\left(\text{x}\right){\text{=a}}_{\text{2k}}{\text{x}}^{\text{2k}}{\text{+a}}_{\text{2k-1}}{\text{x}}^{\text{2k-1}}\text{+}\mathrm{......}{\text{+a}}_{\text{1}}{\text{x+a}}_{\text{0}}$ .

The function can be expressed in the form of:

  $\Rightarrow {\text{a}}_{\text{2k}}{\text{x}}^{\text{2k}}{\text{-a}}_{\text{2k-1}}{\text{x}}^{\text{2k-1}}\text{+}\mathrm{......}{\text{-a}}_{\text{1}}{\text{x+a}}_{\text{0}}$ .

For a function to be even only if:

  $\text{f}\left(\text{-x}\right)\text{=f}\left(\text{x}\right)$ .

The even degree polynomial function is even only if it has:

  ${\text{a}}_{\text{2k-1}}{\text{=a}}_{\text{2k-3}}\text{=}\mathrm{.......}{\text{=a}}_{\text{3}}{\text{=a}}_{\text{1}}\text{=0}$ .

For example:

  $\text{f}\left(\text{x}\right){\text{=x}}^{\text{6}}{\text{-3x}}^{\text{4}}{\text{+x}}^{\text{2}}\text{-1}$ is even.

  $\text{f}\left(\text{x}\right){\text{=x}}^{\text{6}}{\text{+2x}}^{\text{5}}{\text{-x}}^{\text{2}}\text{+x+1}$ .

Hence, every even degree polynomial not even function.