An elegant way to find the range of functions? Find the range of the function  $ y=$ $\Large\frac {x}{x^2+1}$. My idea was to express the value of $x$ in terms of $y$. But that yields a very messy solution involving the quadratic formula. Is there any elegant way to find the range if the use of calculus is allowed?
 A: You can find the extreme values by using derivatives but since you already found the following bounds, you just need to show that these belong to the range; continuity does the rest.

From $(x-1)^2 \ge 0 \implies x^2+1 \ge 2x$, you have:
$$\frac{x}{x^2+1} \le \frac{1}{2}$$
and from $(x+1)^2 \ge 0 \implies x^2+1 \ge -2x$, you have: 
$$-\frac{1}{2} \le \frac{x}{x^2+1}  $$
Now note that:


*

*$f(1) = \tfrac{1}{2}$ and $f(-1) = -\tfrac{1}{2}$;

*$f$ is continuous, so it takes all values between $-\tfrac{1}{2}$ and $\tfrac{1}{2}$.

A: Hint:
1) Note that the function $y=\frac{x}{x^2+1}$ is continuous in the domain $(-\infty,+\infty)$. So the range is the interval between the  absolute minimum and the absolute maximum of the function.
2)Note that the limits at $\pm \infty$ of the function are $0$.
3) take the derivative and search the minimum and maximum  values of the function.
A: The function $f(x)=\cfrac{x}{x^2+1}$ is odd, so it suffices to determine the range for $x \ge 0\,$, then add its symmetric across $0$.
For $x \gt 0\,$, the well known (and easy to prove e.g. by AM-GM) inequality $x+\cfrac{1}{x} \ge 2$ gives $f(x)=\cfrac{1}{x+\frac{1}{x}} \le \cfrac{1}{2}$ with equality at $x=1$. Since $f(0)=0$ and $f$ is continuous, it follows that the image of $[0,\infty)$ is $\left[0,\cfrac{1}{2}\right]$ and, since $f$ is odd, its range is $\left[0,\cfrac{1}{2}\right] \cup \left[-\cfrac{1}{2},0\right] = \left[-\,\cfrac{1}{2},\cfrac{1}{2}\right]\,$.
A: $y=\frac{x}{1+x^2}\implies yx^2-x+y=0$
.
Now Use Shreedharacharya's formula to get $x=\frac{1\pm\sqrt{1-4y^2}}{2y}$.
Value of $x$ is meaningful only if $y \neq 0$ & $1-4y^2\ge0\implies y\in [-1/ 2,1/2]$-{0}
& if $x=0$ then $y=0$,so $y=0$ is in the range.Hence $-1/2\leq y \leq 1/2 $.
A: I don't think it is so messy express $x$ as a function of $y$. 
You can write
$$yx^2-x+y=0$$
$1)$ $y=0$ iff $x=0$.
$2)$ If $y\ne 0$ then:
$$x=\frac{1\pm\sqrt{1-4y^2}}{2y}$$
and once $x\in \Bbb R$ we must have:
$$1-4y^2\ge 0 \to y\in \left[-\frac{1}{2},0\right[\cup \left]0,\frac{1}{2}\right]$$
once we know that $0$ is in the range then it is
$$\left[-\frac{1}{2},\frac{1}{2}\right]$$
A: Show that $$-\frac{1}{2}\le\frac{x}{x^2+1}\le \frac{1}{2}$$
