Why $y=0$ is included in the range of $f(x)=\frac{2x}{x^2 + 1}$?

Question

I have this statement:

Find the range of $$\frac{2x}{x^2 + 1} = f(x)$$

I know that to get the range, I need to write the function in terms of $x$, that is:

$yx^2 - 2x+y=0 \to x = \frac{-1 \pm\sqrt{-y^2 + 1}}{y}$, then the solutions are $(-y^2 +1) \geq 0$ and $y \neq 0$, therefore the range is: $[-1, 0) \cup(0, 1].$ But according to the guide it’s equal to $[-1, 1]$, and why include the $0$ ? I think that if $y = 0$, the expression is indeterminate. So, what is wrong with my development?

Answer 1

Well, $0$ is in the range, $f(0) = 0$, so something is missing in your argument.

Do you recall in proving the quadratic formula, you divide both sides of the equation by the coefficient of the quadratic term? Division by zero is undefined. Normally, we apply the quadratic formula to actual quadratic polynomials. However, when the leading coefficient is a variable, it can be zero or not depending on the value of the variable.

When you do the division in the quadratic formula with an expression that can be zero, you actually get a piecewise result: $$ x = \begin{cases} \frac{-1 \pm\sqrt{-y^2 + 1}}{y} ,& y \neq 0 \\ \frac{-y}{-2} ,& y = 0 \end{cases} \text{.} $$ Of course, the second case simplifies to $x = 0$ when $y = 0$.

Answer 2

$0$ is in the range, because $f(0)=0.$

When you applied the quadratic formula, you assumed $y\ne0.$

You should have also considered the case $y=0$ (separately):

$yx^2-2x+y=0$ when $y=0$ leads to a solution $x=0$.

Answer 3

Note that $(x−1)^2=x^2−2x+1\ge0,$ therefore $$\dfrac{2x}{x^2+1}\le1$$ and this maximum achieve at $x=1.$ The function $f(x)=\dfrac{2x}{x^2+1}$is odd, that is $$f(-x)=-f(x)$$ and therefore the graph of $y=f(x)$ is symmetric in opposite quadrants. Hence the (absolute) minimum of $f$ must occurs at $x=−1.$ Moreover $f(x)\to0$ as $x\to\pm\infty$ and, this gives you a complete picture of its behavior.

For a different approach of the range of $f,$ observe that for any real number $x$ there is a unique real number $\theta\in\left(-\dfrac{\pi}2, \dfrac{\pi}2\right)$ such that $x=\tan\theta.$ Then with a bit of trigonometry we get $$\dfrac{2x}{x^2+1}=\sin 2\theta.$$ Hence $-1\le f(x)\le 1$ is the range of $f.$ For $\theta=0,$ you get that $y=0.$

Since we know that $f$ is odd, it is enough to study it for all $x\gt0.$ If you know hyperbolic functions, you can substitute $x=e^t$ and obtain $y=\text{sech } t$ parameterization. This gives you another way to study the same function.