Question

The minimum of $f\left(x\right)=(x-2{)}^{\frac{2}{3}}$ occurs at $x=$

• A. $2$
• B. $1$
• C. $0$
• D. $-2$

Answer 1

The correct option is A $2$

Given, $f\left(x\right)=(x-2{)}^{\frac{2}{3}}$

$f^{\prime }\left(x\right)=\frac{2}{3}(x-2{)}^{\frac{-1}{3}}$
$f^{\prime }\left(x\right)<0$ when $x<2$
$f^{\prime }\left(x\right)>0$ when $x>2$
So, $f\left(x\right)$ is minimum at $x=2$.