Determining range of a function

Question

I was trying to determine the domain and range of a function. The function is: $$y = \frac{x}{x^2 + 1}$$ I determined the domain which is $\mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$. Then, I tried to determine the range by this equation: $$x = \frac{1 \pm \sqrt{1 - 4y^2}}{2y}$$ I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is $\{-0.5 \leq y < 0\}$ and $\{0 < y \leq 0.5\}$. Zero cannot be a range because the denominator cannot be zero. But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?

Answer 1

A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in $$y = \frac{x}{x^2 + 1},$$ the first steps could be $$y (x^2 + 1) = x,$$ $$y x^2 + - x + y = 0.$$ Now if $y \neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply: $$ -x = 0, $$ $$ x = 0.$$

Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$ that is, to look for any possible ways that $y \neq 0.$ Your quadratic formula then finds the rest of the range for you.

Answer 2

An idea what to do with this kind of questions: assume $\;a\in\Bbb R\;$ belongs to the function's range, then there exists $\;x\in\Bbb R\;$ (the function's domain) s.t.

$$a=f(x)=\frac x{x^2+1}\implies ax^2-x+a=0\implies \Delta=1-4a^2\ge0\implies |a|\le\frac12$$

and thus you get that it must be that $\;-\frac12\le a\le\frac12\;$ , and the range of the function is waaaay different from what you thought.