The number of critical points of the function \(f(x)=(x-2)^{2 / 3}(2 x+1)\) is [JEE Main 2024, 8 Apr (Shift 1)]

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

\(f'(x)=\frac{2}{3}(x-2)^{-\frac{1}{3}}(2x+1)+(x-2)^{\frac{2}{3}}\)

\(f'(x)=2\times \frac{(2x+1)+(x-2)}{3(x-2)^{\frac{1}{3}}}\)

\(\frac{3x-1}{(x-2)^{\frac{1}{3}}}=0\)

Critical points \(x=\frac{1}{3}\) and \(x=2\)