Question

The critical points of the function $f\left(x\right)={(x-2)}^{2/3}(2x+1)$ are

• A. $-1,2$
• B. $1$
• C. $1,-\frac{1}{2}$
• D. $1,2$

Answer 1

The correct option is D $1,2$

Given, $f\left(x\right)=(x-2{)}^{2/3}(3x+1)$

$f^{\prime }\left(x\right)=\frac{2}{3}{(x-2)}^{\frac{-1}{3}}(2x+1)+2{(x-2)}^{\frac{2}{3}}=10(x-1){(x-2)}^{\frac{-1}{3}}$
$f^{\prime }\left(x\right)=0$

$\Rightarrow x=1$
Also $f^{\prime }\left(x\right)$ does not exits at $x=2$
Hence the critical points are $x=1,2$