QuestionDetermine if the function $g(x)=x^{3}-7x$ is even, odd, or neither, and describe its symmetry.

Studdy Solution

STEP 1

Assumptions1. The function given is $g(x)=x^{3}-7 x$ . A function is even if $f(-x) = f(x)$ for all $x$ in the domain of $f$
3. A function is odd if $f(-x) = -f(x)$ for all $x$ in the domain of $f$
4. If a function is neither even nor odd, it does not satisfy either of the above conditions

STEP 2

To determine if the function is even, odd, or neither, we will first calculate $g(-x)$ .
$g(-x) = (-x)^{}-7 (-x)$

STEP 3

implify the expression.
$g(-x) = -x^{3}+7 x$

STEP 4

Now, we compare $g(-x)$ with $g(x)$ and $-g(x)$ .
The function $g(x) = x^{3}-7 x$ and $g(-x) = -x^{3}+7 x$ are not equal, so the function is not even.
The function $-g(x) = -(x^{3}-7 x) = -x^{3}+7 x$ and $g(-x) = -x^{3}+7 x$ are equal, so the function is odd.

STEP 5

The symmetry of an odd function is origin symmetry. This means that the function is symmetric about the origin.
The function $g(x)=x^{3}-7 x$ is an odd function and has origin symmetry.

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