Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Slaodsc We have $f(x)=(x-2{)}^{\frac{2}{3}}(2x+1)$
$\therefore f^{\prime }(x)=2(x-2{)}^{\frac{2}{3}}+\frac{2}{3}(2x+1)(x-2{)}^{\frac{-1}{3}}=\frac{2(3(x-2)+2x+1)}{3(x-2{)}^{\frac{1}{3}}}=\frac{2(5x-5)}{3(x-2{)}^{\frac{1}{3}}}=\frac{10(x-1)}{3(x-2{)}^{\frac{1}{3}}}$
As domain of $f$ is $R$, so $x=1$ and $x=2$ are two critical points of $f(x)$.