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College Algebra : Symmetry

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Example Question #11 : Symmetry

Consider the function $f(x)= x^{8}+ 2x^{4}+ 1$ $\displaystyle f(x)= x^{8}+ 2x^{4}+ 1$ .

Is $\displaystyle f$ an even function, an odd function, or neither?

Possible Answers:

Even

Neither

Odd

Correct answer:

Even

Explanation:

A function $\displaystyle f$ is even if, for each $\displaystyle c$ in its domain,

$\displaystyle f(-c)= f(c)$ .

It is odd if, for each $\displaystyle c$ in its domain,

$\displaystyle f(-c)=- f(c)$ .

Substitute $\displaystyle -x$ for $\displaystyle x$ in the definition:

$f(-x)= (-x)^{8}+ 2(-x)^{4}+ 1$ $\displaystyle f(-x)= (-x)^{8}+ 2(-x)^{4}+ 1$

$= (-1)^{8}x^{8}+ 2 (-1)^{4}x^{4}+ 1$ $\displaystyle = (-1)^{8}x^{8}+ 2 (-1)^{4}x^{4}+ 1$

$= 1x^{8}+ 2 (1) x^{4}+ 1$ $\displaystyle = 1x^{8}+ 2 (1) x^{4}+ 1$

$= x^{8}+ 2x^{4}+ 1$ $\displaystyle = x^{8}+ 2x^{4}+ 1$

= f(x) $\displaystyle = f(x)$

Since $\displaystyle f(x) = f(-x)$ , $\displaystyle f$ is an even function.

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Example Question #12 : Symmetry

f(x) $\displaystyle f(x)$ is an even function. Let $g(x) = \frac{f(x)}{x}$ $\displaystyle g(x) = \frac{f(x)}{x}$ .

Is $\displaystyle g$ an even function, an odd function, or neither?

Possible Answers:

Even

Odd

Neither

Correct answer:

Odd

Explanation:

A function $\displaystyle f$ is even if, for each $\displaystyle c$ in its domain,

$\displaystyle f(-c)= f(c)$ .

It is odd if, for each $\displaystyle c$ in its domain,

$\displaystyle f(-c)=- f(c)$ .

Substitute $\displaystyle -x$ for $\displaystyle x$ in the definition of $\displaystyle g$ :

$g(-x) = \frac{f(-x)}{-x} = - \frac{f(-x)}{x}$ $\displaystyle g(-x) = \frac{f(-x)}{-x} = - \frac{f(-x)}{x}$

Since $\displaystyle f$ is even, $\displaystyle f(-x)= f(x)$ , so

$g(-x) = - \frac{f(x)}{x} = -g(x )$ $\displaystyle g(-x) = - \frac{f(x)}{x} = -g(x )$

This makes $\displaystyle g$ an odd function.

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College Algebra : Symmetry

Study concepts, example questions & explanations for College Algebra

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Example Question #11 : Symmetry

Example Question #12 : Symmetry

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