Question

Number of critical points of the function $f\left(x\right)={(x-2)}^{\frac{2}{3}}(2x+1)$ is equal to

Answer 1

$f\left(x\right)=(x-2{)}^{\frac{2}{3}}(2x+1)$
$f^{\prime }\left(x\right)=\frac{2}{3}{(x-2)}^{-\frac{1}{3}}(2x+1)+{(x-2)}^{\frac{2}{3}}\left(2\right)$
$=\frac{2(\frac{2x+1}{3}+(x-2\left)\right)}{{(x-2)}^{\frac{1}{3}}}$
For critical point $f^{\prime }\left(x\right)=0or{f}^{\prime }\left(x\right)$ is not defined so $x=2and2x+1+3x-6=0$
$\Rightarrow x=1$
Therefore, number of critical points are $2$