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When is an equation of the form $ y = f(x)$ not a function? Does it need to be surjective, injective, both, neither? I vaguely remember something called the vertical line test, but it's not making any connections in my brain.

Also, I'm reading a book about differential equations and there's a sentence in it that brought this question to mind. It reads, "the equation $ y = \sqrt{-(1+x^2)}$ does not define a function." What is it about that equation/corresponding graph that fails the definition of a function?

HeroZhang001
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DBSHomie
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    This seems to be well addressed by the wikipedia page. Is there a point there that is unclear? – Noah Schweber Feb 05 '23 at 01:24
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    Just to add: your book probably means that $x \mapsto \sqrt{-1 - x^2}$ doesn't define a real-valued function. Notice that $-1 - x^2$ is always negative and so it doesn't have a real square root. – Joseph Harrison Feb 05 '23 at 01:28
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    If one was feeling a bit mischievous, they could argue that your last equation does define a function, but one whose domain is the empty set. But that's a rather pathological example of a function. – JonathanZ Feb 05 '23 at 01:38
  • consult any precalculus textbook, they elaborate a bit more on this topic. – CroCo Feb 05 '23 at 07:00

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For a function $y=f(x)$ to be called a 'function' it has to meet a certain criteria,

(i) There should be a value of $y$ for every value of $x$

(ii) There should be only one value of $y$ for every value of $x$(This is why we do vertical line test)

(iii) For every value of $x$, $y$ should be defined.

Vertical line test:- If any vertical line cuts the graph of the given function at more than one point, then it's not a function. It means for a single value of $x$, if you see the graph giving more than one value of $y$ then it's not a function (as stated above)

1. Let's take $y=x(x-1)(x-2)$

Notice that any vertical line does not cut the curve at more than one point, also notice that for every value of $x$ in its domain($x\in R$), $y$ is defined(the curve is continuous everywhere in its domain).

So $y=x(x-1)(x+1)$ is a 'function'

2. Now let's say $y^2=1-x^2$

Notice that any vertical line cuts the curve at more than one point, that means for a single value of $x$, we get more than one value of $y$.

So $y^2=1-x^2$ is not a 'function'

Now your question about $y=\sqrt{-(1+x^2)}$

$1+x^2$ is always $\ge 1$ for any value of $x$

So, $-(1+x^2)$ is always $\le -1$ for any value of $x$

That means the thing inside the square root always remains negative for any value of $x$. So, there is no real value of $y$ for any value of $x$. That's why it's not a function.

I hope its clear now.

Adil Ahamed
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