Range of a 1-2 function

Question

$$f(x)=\frac x {x^2+1}$$ I want to find range of $f(x)$ and I do like below . If someone has different Idea please Hint me . Thanks in advanced .

This is 1-1 function $\\f(x)=\dfrac{ax+b}{cx+d}\\$, This is 2-2 function $\\f(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'}\\$, This is 1-2 function $\\f(x)=\dfrac{ax+b}{a'x^2+b'x+c'}\\$

Answer 1

Method 5

$0\le(x+1)^2 = (x^2+1)+2x$ dividing by $(x^2+1)$ we have $0\le1+2y$ $0\le(x-1)^2 = (x^2+1)-2x$ dividing by $(x^2+1)$ we have $0\le1-2y$

And $y\in[-\frac12,\frac12]$.

Answer 2

Method 1

$$y=\dfrac{x}{x^2+1} \to yx^2-x+y=0 \\\Delta \geq 0 \to 1-4(y)(y) \geq0 \\y^2 \leq \dfrac14 \\ y \in \left[-\dfrac12,\dfrac12\right]$$

Method 2

It is well known that$$\left|x+\frac1x\right| \geq 2 \to y=\dfrac{x}{x^2+1} \to \dfrac1y=\dfrac{x^2+1}{x}=x+\dfrac1x \\\to |\dfrac1y|\geq 2 \\|y| \leq \dfrac12 \space y \in\left[-\dfrac12,\dfrac12\right]$$

Method 3

$$y=\frac{x}{x^2+1} \to y'=\frac{1-x^2}{(x^2+1)^2}=0 \to x=\pm1 \\ \begin{cases}x=1 & f(1)=\frac12 \\x=-1 & f(1)=-\dfrac12\\ x\to -\infty & \lim_{x \rightarrow -\infty}f(x) \to 0\\x\to -\infty & \lim_{x \rightarrow +\infty}f(x) \to 0\end{cases} \to y \in\left[-\frac12, \dfrac12 \right] $$

Method 4

By substituting $x=\tan \alpha$ we have $$y=\dfrac x {x^2+1}=y=\frac{\tan \alpha}{\tan^2 \alpha+1}=\frac{\dfrac{\sin \alpha}{\cos \alpha}}{\left(\dfrac{\sin \alpha}{\cos \alpha}\right)^2+1}=\\=\sin \alpha \cdot \cos \alpha=\frac22 \sin \alpha \cdot\cos \alpha =\frac12 \sin 2\alpha \\ -1 \leq \sin 2\alpha \leq 1 \\ \to -\frac12 \leq \frac12 \sin 2\alpha \leq \frac12 \to y \in \left[-\dfrac12,\dfrac12\right]$$