Question

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

• A. $-1$ and $2$
• B. $1$
• C. $1$ and $-2$
• D. $1$ and $2$

Answer 1

Answer: (B)

$1$

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

$\Rightarrow f^{\prime }\left(x\right)=\frac{2}{3}(x-2{)}^{-1/3}(2x+1)+(x-2{)}^{2/3}\cdot 2$

$=\frac{2(2x+1)}{3(x-2{)}^{1/3}}+2(x-2{)}^{2/3}$

$=\frac{2(2x+1)+6(x-2)}{3(x-2{)}^{1/3}}$

$=\frac{4x+2+6x-12}{3(x-2{)}^{1/3}}$

$=\frac{10(x-1)}{3(x-2{)}^{1/3}}$

For critical points,

$f^{\prime }\left(x\right)=0$

$\Rightarrow x=1$

and $f^{\prime }\left(x\right)$ is not defined at x= 2.