PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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F(X) = (X - 1) (X + 2)2. - PUC Karnataka Science Class 12 - Mathematics

Question

f(x) = (x - 1) (x + 2)2.

Solution

$\text { Given }: \hspace{0.167em} f\left( x \right) = \left( x - 1 \right) \left( x + 2 \right)^2$

$= \left( x - 1 \right)\left( x^2 + 4x + 4 \right)$

$= x^3 + 4 x^2 + 4x - x^2 - 4x - 4$

$= x^3 + 3 x^2 - 4$

$\Rightarrow f'\left( x \right) = 3 x^2 + 6x$

$\text { For the local maxima or minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow 3 x^2 + 6x = 0$

$\Rightarrow 3x\left( x + 2 \right) = 0$

$\Rightarrow x = 0 \text { and } - 2$

$\text { Thus, x = 0 and x = - 2 are the possible points of local maxima or local minima }.$

$\text { Now },$

$f''\left( x \right) = 6x + 6$

$\text { At x } = 0:$

$f''\left( 0 \right) = 6\left( 0 \right) + 6 = 6 > 0$

$\text { So, x = 0 is the point of local minimum } .$

$\text { The local minimum value is given by }$

$f\left( 0 \right) = \left( 0 - 1 \right) \left( 0 + 2 \right)^2 = - 4$

$\text { At }x = - 2:$

$f''\left( - 2 \right) = 6\left( - 2 \right) + 6 = - 6 < 0$

$\text { So, x = - 2 is the point of local maximum} .$

$\text { The local maximum value is given by }$

$f\left( - 2 \right) = \left( - 2 - 1 \right) \left( - 2 + 2 \right)^2 = 0$

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Solution F(X) = (X - 1) (X + 2)2. Concept: Graph of Maxima and Minima.
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