# Is f(x) = x^4-64 an odd function or an even function?

Feb 11, 2017

$f \left(x\right) = {x}^{4} - 64$ is even.

#### Explanation:

An function $f \left(x\right)$ is even if $f \left(\text{-} x\right) = f \left(x\right)$. What does this mean? It means that for any given input $x$, the function $f$ does the same thing to $x$ as it does to $- x$. When we give it $x$, it gives us $y$ back. When we give it $- x$, it gives us $y$ again.

The visual interpretation to this is that an even function is symmetrical about the $y$-axis. It has to be, because no matter what $x$ we choose, the $y$-value for $x$ matches the $y$-value for $- x$.

To see if a function $f$ is even, we ask: when $f$ gets the input $- x$, does it return the same output as if we had given it $x$? In math terms, we're asking:

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

For the given function $f \left(x\right) = {x}^{4} - 64$, $f$ takes an input, computes its 4th power, then subtracts 64. To test if $f$ is even, we plug $- x$ in as our input and see if we still get ${x}^{4} - 64$ back.

We compute:

$f {\left(\text{-"x)=("-} x\right)}^{4} - 64$
color(white)(f("-"x))=("-"1)^4(x)^4 -64
$\textcolor{w h i t e}{f \left(\text{-} x\right)} = 1 \times {x}^{4} - 64$
$\textcolor{w h i t e}{f \left(\text{-} x\right)} = {x}^{4} - 64$

And hey look: $f \left(\text{-} x\right) = {x}^{4} - 64$, which is $f \left(x\right)$! Thus, $f \left(\text{-} x\right) = f \left(x\right)$, and so $f$ is even.