Question

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

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

Answer 1

Answer: (A)

$f\left(x\right)={\left(x-2\right)}^{2/3}\left(2x+1\right)$
$f^{\prime }\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{-1/3}\left(2x+1\right)+{\left(x-2\right)}^{2/3}\cdot 2$
Clearly $f^{\prime }\left(x\right)$ is not defined at $x=2$
$\therefore x=2$ is a critical point.
Another critical point is given by $f^{\prime }\left(x\right)=0$
i.e., $\frac{2}{3}\frac{2x+1}{{\left(x-2\right)}^{1/3}}+2{\left(x-2\right)}^{2/3}=0$
$\Rightarrow \frac{2}{3}\left(2x+1\right)+2\left(x-2\right)=0$
$\Rightarrow 4x+2+6x-12=0$
$\Rightarrow 10x-10=0$
$\Rightarrow x=0$
Hence critical points are $1$ and $2$.