Question #0dd78

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Question #0dd78

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Answer 1 · 2018-01-08T18:54:37

$g (x)$ $g(x)$ is an even function.

Explanation:

If $g (- x) = g (x)$ $g(-x)=g(x)$ , the function is even, and if $g (- x) = - g (x)$ $g(-x)=-g(x)$ , it is odd.

To find out which one it is, we plug in $- x$ $-x$ :
$g (- x) = {(- x)}^{2} + {cos}^{2} (- x)$ $g(-x)=(-x)^2+cos^2(-x)$

Cosine is an even function, and a negative squared is positive, so the minus signs just disappear:
$g (- x) = x^{2} + {cos}^{2} (x) = g (x)$ $g(-x)=x^2+cos^2(x)=g(x)$

Since $g (- x) = g (x)$ $g(-x)=g(x)$ , we can conclude that $g (x)$ $g(x)$ is an even function.

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Question #0dd78

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