Inspired by Rudi Mentar

@Vincent Martin – Wrong. Division by zero is undefined. Taking the square root of a negative number is not the same as this.

@Yasharyan Gaikwad – cough cough

@Kenneth Choo – And what matters if $\sqrt{x}$ cannot be negative? $f(\sqrt{x})=x^2$ means " $f$ evaluated in $\sqrt{x}$ is $x^2$ when the expressions are defined". It does not say anything else. It does not forbid anyone to take the value of $f(\sin x$ ).

Taking $f(x)=x^4,g(x)=\sqrt{x}, h(x)=x^2$ , with domains $\mathbb{R},\mathbb{R^+\cup 0},\mathbb{R^+\cup 0}$ respectively satisfies the equation $f\circ g=h$ , because the value of $f$ when $x<0$ will never be used.

The problem is that you're thinking of $f$ as " $f$ restricted to the domain of $g$ " , and that is not necessarily the case.

When you talk about how it satisfies the equation you're talking about the domain of the composite function $f\circ g$ , not the domain of $f$ . I will mention something I mentioned above.

This kind of problem is very common in all mathematics. For example, the Riemann $\zeta$ function is a function such that $\zeta(\sqrt{x}+1)=\sum_{n=1}^\infty \frac{1}{n^{\sqrt{x}+1}},$ for all nonnegative real number $x$ , and that is analytic in the complex plane.

And $\zeta(-2)=0$ .

http://www.wolframalpha.com/input/?i=Riemann+zeta(-2) https://en.wikipedia.org/wiki/Particular values of Riemann zeta_function

The picture shows a partial sum of the mentioned series, and the function $\zeta(\sqrt{x}+1)$ . In the case of $x=0$ the series becomes the Harmonic series, and so diverges. The function also goes to infinity there. It's grafically clear that the series and the Zeta function evaluated in $\sqrt{x}+1$ are the same for all nonnegative $x$ .

In fact we do these kinds of continuations of functions more often. In linear algebra when we want a linear mapping we define it for the basis and "extend it by linearity". In group theory we define it for the generators of a group, and "extend by homomorphism property". The same we do with rings. In complex analysis we take a function defined in an open, connected set of the complex plane (In this case, the complex with real part greater than 1), which is infinitely differentiable there, and extend it to the whole complex plane. In Fourier Analysis we take a bounded (piecewise continuous) function defined on an interval $(0,\pi]$ and extend it by parity to [-\pi,\pi] making it odd or even, depending of what we want, to make a Fourier sine or cosine expansion of the function.

Going back to this case, here we define a function for the nonnegative integers (i.e. The real numbers of the form $\sqrt{x}$ ) and see that is not enough to extend it to all real numbers. In particular, without any restriction $f(-2)$ could be any real number.

You are asuming that if $f(z)$ is ever defined, then $z=\sqrt{x}$ for some $x$ , and that doesn't have to be true. If you want to use that, then you will have to prove it . You'll have to prove that thre isn't any function $f$ defined for all real numbers, such that $f(\sqrt{x})=x^2$ for all nonnegative real numbers, using, of course, the definition of function and it's basic theorems .

@Kenneth Choo – The solution of the problem cannot be determined because there are infinite solutions. $f$ may be defined as $x^4$ for nonnegative real numbers, and whatever for negative real numbers. And if $g(x)=\sqrt{x}$ , then the composite $f\circ g (x)$ will always equal $x^2$ , since the domain of $f\circ g$ is the same domain of $g$ (Nonnegative real numbers), even if the domain of $f$ is the quaternions (If we take $f(x)=x^4$ that could be the case). Let $\mathbb{R}^o=\mathbb{R}^+\cup\{0\}$ . We may have (taking $f(x)=x^4,g(x)=\sqrt{x},h(x)=x^2$ ):

$f:\mathbb{H}\to\mathbb{H}$

$g:\mathbb{R}^o\to\mathbb{R}^o\subseteq \mathbb{H}$

$f\circ g: \mathbb{R}^o\to \mathbb{H}$

And $f\circ g=h$ .

All of this independently of the domain of $f$ . And of the value of $f$ in $\mathbb{H}-\mathbb{R}^o$ .

@Kenneth Choo – I can say -2 has no relation to sqrt(x) because of the lack of information telling me that I need to view that as a restriction on my domain. I see nothing telling me that it's a parametric or a composite. Therefore when I look at it, f(sqrt(x)) doesn't mean "the domain of x must be non-negative." Given the lack of extra information, the ONLY way I can view it is that the sqrt(x) and the x^2 is to ultimately establish that f(x) = x^4.

I will accept "undefined" only with more specific info. Otherwise the "undefined" answer is just trying to be too clever.

@Kenneth Choo – Nope. You are Trying to break the definition of function.