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Question
f(x) = (x \[-\] 1) (x \[-\] 2)2.
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Solution
\[\text { Given: } f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right)^2 \]
\[ = \left( x - 1 \right)\left( x^2 - 4x + 4 \right)\]
\[ = x^3 - 4 x^2 + 4x - x^2 + 4x - 4\]
\[ = x^3 - 5 x^2 + 8x - 4\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 - 10x + 8\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 3 x^2 - 10x + 8 = 0\]
\[ \Rightarrow 3 x^2 - 6x - 4x + 8 = 0\]
\[ \Rightarrow \left( x - 2 \right)\left( 3x - 4 \right) = 0\]
\[ \Rightarrow x = 2 \text { and }\frac{4}{3}\]
\[\text { Thus, x = 2 and } x = \frac{4}{3} \text { are the possible points of local maxima or local minima } . \]
\[\text { Now }, \]
\[f''\left( x \right) = 6x - 10\]
\[At x = 2: \]
\[ f''\left( 2 \right) = 6\left( 2 \right) - 10 = 2 > 0\]
\[\text { So, x = 2 is the point of local minimum }. \]
\[\text { The local minimum value is given by }\]
\[f\left( 2 \right) = \left( 2 - 1 \right) \left( 2 - 2 \right)^2 = 0\]
\[\text { At }x = \frac{4}{3}: \]
\[ f''\left( \frac{4}{3} \right) = 6\left( \frac{4}{3} \right) - 10 = - 2 < 0\]
\[\text { So, x} = \frac{4}{3}\text { is the point of local maximum } . \]
\[\text { The local maximum value is given by }\]
\[f\left( \frac{4}{3} \right) = \left( \frac{4}{3} - 1 \right) \left( \frac{4}{3} - 2 \right)^2 = \frac{1}{3} \times \frac{4}{9} = \frac{4}{27}\]
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