Question

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

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (D)

2

$f^{\prime }\left(x\right)=1-\frac{4}{(x+2{)}^2}$
$=\frac{(x+2{)}^2-2^2}{(x+2{)}^2}$
$f^{\prime }\left(x\right)=\frac{(x+4)x}{(x+2{)}^2}$
$f^{\prime }\left(x\right)=0$ at $x=0,-4$
$f^{\prime \prime }\left(x\right)=\frac{8}{(x+2{)}^3}$
$f^{\prime \prime }\left(x\right)>0$ at $x=0$
$f\left(x\right)$ has minimum at $x=0$
$f\left(0\right)=2$