Question

Find the critical points of the function $f(x)={(x-2)}^{2/3}(2x+1)$ .

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

Answer 1

Answer: (B)

We have, $f(x)={(x-2)}^{3/2}(2x+1)$
$\Rightarrow$ $f^{\prime }(x)=\frac{2}{3}{(x-2)}^{-1/3}(2x+1)$
$+{(x-2)}^{2/3}\cdot 2$
$=\frac{2(2x+1)}{3{(x-2)}^{1/3}}+2{(x-2)}^{2/3}$
$=\frac{2(2x+1)+6(x-2)}{3{(x-2)}^{1/3}}$
$=\frac{4x+2+6x-12}{3{(x-2)}^{1/3}}$
$=\frac{10(x-1)}{3{(x-2)}^{1/3}}$
For critical points, $f^{\prime }(x)=0$
$\Rightarrow$ $x=1$ and $f^{\prime }(x)$ is not defined at $x=2.$