Is $\sqrt{x}$ an even function

Question

I'm going through some pre-calculus and I am presented with this rule.

If $f(x)$ is an even function, then $f(x) = f(-x)$

So with the following example I have:

$$f(x) = 3x^4 \\ f(-x) = 3(-x)^4 = 3x^4$$

However, when I look at the graph of $f(x) = \sqrt{-x}$ it is the reflection of $f(x) = \sqrt{x}$ about the y axis, but I cannot get this to work out algebraically...

$$f(x) = \sqrt{x} = f(-x) = \sqrt{-x}$$ which is is true, because of domain restrictions, but $f(x)$ does not look like $f(-x)$ like in the previous example. Is it because the domain changes for $\sqrt{-x}$ that I can do some trick to make turn it into $\sqrt{x}$ to allow $f(x) = f(-x)$?

For instance,

But $Dom(f(-x)) = (-\infty, 0]$, therefore $-x = x$?? I'm not sure how to finish this off algebraically? Help anyone?

Answer 1

If $f(x)$ is a even function in $\mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.

Answer 2

In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But $$f(-x)=\sqrt{-x}=\pm\sqrt{x}i=if(x)$$ and $\pm i\neq 1$, so clearly it's not even.