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Q.
Number of critical points of the function $f(x)=(x-2)^{\frac{2}{3}}(2 x+1)$ is equal to
Application of Derivatives
Solution:
Slaodsc We have $f(x)=(x-2)^{\frac{2}{3}}(2 x+1)$
$\therefore f ^{\prime}( x )=2( x -2)^{\frac{2}{3}}+\frac{2}{3}(2 x +1)( x -2)^{\frac{-1}{3}}=\frac{2(3( x -2)+2 x +1)}{3( x -2)^{\frac{1}{3}}}=\frac{2(5 x -5)}{3( x -2)^{\frac{1}{3}}}=\frac{10( x -1)}{3( x -2)^{\frac{1}{3}}}$
As domain of $f$ is $R$, so $x =1$ and $x =2$ are two critical points of $f ( x )$.