# Is the function f(x) = x^3 symmetric with respect to the y-axis?

Mar 18, 2018

No, it has rotational symmetry of order $2$ about the origin.

#### Explanation:

• An even function is a function satisfying:
$f \left(- x\right) = f \left(x\right) \text{ }$ for all $x$ in the domain of $f \left(x\right) \textcolor{w h i t e}{\frac{0}{0}}$

• An odd function is a function satisfying:
$f \left(- x\right) = - f \left(x\right) \text{ }$ for all $x$ in the domain of $f \left(x\right) \textcolor{w h i t e}{\frac{0}{0}}$

Even functions are symmetric with respect to the $y$-axis.

Odd functions have rotational symmetry of order $2$ about the origin.

Given:

$f \left(x\right) = {x}^{3}$

Note that for any value of $x$:

$f \left(- x\right) = {\left(- x\right)}^{3} = {\left(- 1\right)}^{3} {x}^{3} = - {x}^{3} = - f \left(x\right)$

So $f \left(x\right) = {x}^{3}$ is an odd function.

It is not symmetric with respect to the $y$-axis, but it has rotational symmetry of order $2$ about the origin.

graph{x^3 [-5, 5, -10, 10]}

In fact any polynomial consisting of only terms of odd degree will be an odd function and any polynomial consisting of only terms of even degree will be an even function.