Is $\sqrt{x}$ an even function?

From my Calculus class I don't think that I would say that the function $f:[0.\infty) \to \mathbb{R}$ given by $f(x) = \sqrt{x}$ is an even function. The graph isn't symmetric about the $y$-axis.

But according to my book, and to Wikipedia a function $f$ is even if $f(-x) = f(x)$ for all $x$ and $-x$ in the domain of $f$. So since the domain of the square root function is $[0,\infty)$, the function would satisfy this.

Is that correct?

Solution 1:

I think that by reading carefully and correctly you have found a subtle problem with the wikipedia definition. It is indeed true that "whenever both $x$ and $-x$ are in the domain of the square root function the function values agree" because (as you know quite well) only $x=0$ satisfies that hypothesis. So according to the strict reading of the definition, the square root function is even.

But (as you also realize) that's not the intent of the definition. Implicit in that definition is the assumption that the domain is symmetric about $0$, so that it contains $-x$ whenever it contains $x$.

I disagree with most (but by no means all) of the other answers and comments. They are just reminding you that the domain is $[0, \infty)$.

Good for you for paying this kind of attention.

Solution 2:

The language used at that Wikipedia article is innacurate. A good definition would be: if $D\subset\mathbb R$, a function $f\colon D\longrightarrow\mathbb R$ is even if $(\forall x\in D):-x\in D\text{ and }f(-x)=f(x)$. Under this definition, I hope that you agree that the square root function (or, for that matter, any function whose domain is $[0,+\infty)$) is not even.