Question

The minimum value of $f\left(x\right)=x+\frac{4}{x+2},x\ge 0$ is

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$f\left(x\right)=x+\frac{4}{x+2}$
$\Rightarrow f^{\prime }\left(x\right)=1-\frac{4}{(x+2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x(x+4)}{(x+2{)}^2}\ge 0,\forall x>0$
So $f\left(x\right)$ is increasing function
$\Rightarrow f\left(x\right)\ge f\left(0\right)$
So, $x=0$ is the only critical point
$f\left(0\right)=2$
when $x\to \infty ,f\to \infty$
$\therefore$ The minimum value of $f\left(x\right)$ is $2$