Question

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

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

Answer 1

Answer: (B)

We have, $ f(x)={{(x-2)}^{3/2}}(2x+1) $
$ \Rightarrow $ $ f'(x)=\frac{2}{3}{{(x-2)}^{-1/3}}(2x+1) $
$ +{{(x-2)}^{2/3}}\cdot 2 $
$ =\frac{2(2x+1)}{3{{(x-2)}^{1/3}}}+2{{(x-2)}^{2/3}} $
$ =\frac{2(2x+1)+6(x-2)}{3{{(x-2)}^{1/3}}} $
$ =\frac{4x+2+6x-12}{3{{(x-2)}^{1/3}}} $
$ =\frac{10(x-1)}{3{{(x-2)}^{1/3}}} $
For critical points, $ f'(x)=0 $
$ \Rightarrow $ $ x=1 $ and $ f'(x) $ is not defined at $ x=2. $