Question

The range of $f\left(x\right)=\frac{x}{x^2+4}$ is

• A. $[-4,4]$
• B. $[-\frac{1}{2},\frac{1}{2}]$
• C. $[-\frac{1}{5},\frac{1}{5}]$
• D. $[-\frac{1}{4},\frac{1}{4}]$

Answer 1

The correct option is D $[-\frac{1}{4},\frac{1}{4}]$

Let
$f\left(x\right)=y=\frac{x}{x^2+4}\Rightarrow y{x}^2-x+4y=0$

As $x\in R$, so $D\ge 0$
$\Rightarrow 1-16y^2\ge 0\Rightarrow 16y^2-1\le 0\Rightarrow y^2\le \frac{1}{16}\Rightarrow y\in [-\frac{1}{4},\frac{1}{4}]$