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?
 A: 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.
A: I think the wikipedia definition is rescued by this sentence:

The concept of evenness or oddness is defined for functions whose domain and image both have an additive inverse.

The domain $[0,\infty)$ does not have an additive inverse. So the remainder of the article doesn't even apply to it, and it's not surprising that attempting to apply it results in nonsense.
On the other hand, if $D$ is a subset of $\Bbb{R}$ which is symmetric about $0$, we can define an additive inverse on $D$ by restriction from $\Bbb{R}$, so is reasonable to ask whether functions from $D$ to $\Bbb{R}$ are even.
A: If $x = 4$, then it is not true that $f(x) = f(-x)$, because $f(-x)$ is not even defined.  Therefore, $f$ is is not even.  However the function $f(x) = \sqrt{|x|}$ is even.  
A: No because if $x>0$ then $f(-x)$ doesn't even make sense (as a real number). The definition given in Wikipedia is for functions of a real value, i.e, functions from $\mathbb{R}$ to $\mathbb{R}$. In your case that's not satisfied. You can, however, enlarge the definition as follows: "A function $f:D\to\mathbb{R}$ is even if $\forall x\in D,-x\in D\,\land f(-x)=f(x)$". In your case, the domain is not symmetric, so the function is not even.
A: 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.
A: HINT: $f(x)=\sqrt{x}$ is defined for all $x$ with $$0\le x<+\infty$$ and $$\sqrt{-x}=-\sqrt{x}$$ makes no sense and $$\sqrt{-x}=\sqrt{x}$$ also
A: It doesn't matter.
Function parity is useful for functions defined both in the negatives and positives.
Whatever the classification of $\sqrt x$, it is of no use.
