Question

$f\left(x\right)=\frac{x}{x^2+5x+4}$ is maximum at $x=$

• A. $2$
• B. $-2$
• C. $6$
• D. $-6$

Answer 1

The correct option is B $-2$

Given: $f\left(x\right)=\frac{x}{x^2+5x+4}$

For maxima or minima,
$f^{\prime }\left(x\right)=\frac{(x^2+5x+4)\times \left(1\right)-\left(x\right)\times (2x+5)}{(x^2+5x+4{)}^2}=0$

$\Rightarrow x^2+5x+4-2x^2-5x=0$

$\Rightarrow -x^2+4=0$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

For $x=2,$

$f\left(x\right)$ $=\frac{2}{4+10+4}=\frac{2}{18}=\frac{1}{9}$

For $x=-2$

$f\left(x\right)$ $=\frac{-2}{4-10+4}=1$

Hence, $f\left(x\right)$ is maximum at $x=-2$.