PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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F(X) = − ( X − 1 ) 3 ( X + 1 ) 2 . - PUC Karnataka Science Class 12 - Mathematics

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Question

f(x) = \[- (x - 1 )^3 (x + 1 )^2\] .

Solution

\[\text { Given:} f\left( x \right) = - \left( x - 1 \right)^3 \left( x + 1 \right)^2 \]

\[ \Rightarrow f'\left( x \right) = - \left[ 3 \left( x - 1 \right)^2 \left( x + 1 \right)^2 + 2\left( x + 1 \right) \left( x - 1 \right)^3 \right]\]

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

\[ f'\left( x \right) = 0\]

\[ \Rightarrow - 3 \left( x - 1 \right)^2 \left( x + 1 \right)^2 - 2\left( x + 1 \right) \left( x - 1 \right)^3 = 0\]

\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 3\left( x + 1 \right) - 2\left( x - 1 \right) \right] = 0\]

\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 3x - 3 - 2x + 2 \right] = 0\]

\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 5x - 1 \right] = 0\]

\[ \Rightarrow x = 1, - 1 \text { and }\frac{- 1}{5}\]

\[\text { Thus, x = 1, x = - 1 and } x = \frac{- 1}{5} \text { are the possible points of local maxima or local minima }. \]

\[\text { Now,} \]

\[f''\left( x \right) = - \left[ 3\left\{ 2\left( x - 1 \right) \left( x + 1 \right)^2 + 2\left( x + 1 \right) \left( x - 1 \right)^2 \right\} + 2\left\{ \left( x - 1 \right)^3 + 3 \left( x - 1 \right)^2 \left( x + 1 \right) \right\} \right]\]

\[ = - 6\left( x - 1 \right) \left( x + 1 \right)^2 + 6\left( x + 1 \right) \left( x - 1 \right)^2 - 2 \left( x - 1 \right)^3 - 6 \left( x - 1 \right)^2 \left( x + 1 \right)\]

\[\text { At x} = 1: \]

\[ f''\left( 1 \right) = - 6\left( 1 - 1 \right) \left( 1 + 1 \right)^2 + 6\left( 1 + 1 \right) \left( 1 - 1 \right)^2 - 2 \left( 1 - 1 \right)^3 - 6 \left( 1 - 1 \right)^2 \left( 1 + 1 \right) = 0\]

\[\text { So, it is a point of inflexion } . \]

\[\text { At } x = - 1: \]

\[ f''\left( - 1 \right) = - 6\left( - 1 - 1 \right) \left( - 1 + 1 \right)^2 + 6\left( - 1 + 1 \right) \left( - 1 - 1 \right)^2 - 2 \left( - 1 - 1 \right)^3 - 6 \left( - 1 - 1 \right)^2 \left( - 1 + 1 \right) = 16 > 0\]

\[\text{ So, x = - 1 is the point of local minimum }. \]

\[\text { The local minimum value is given by } \]

\[f\left( - 1 \right) = - \left( 1 - 1 \right)^3 \left( - 1 + 1 \right)^2 = 0\]

\[\text { At } x = - \frac{1}{5}: \]

\[ f''\left( - \frac{1}{5} \right) = - 6\left( - \frac{1}{5} - 1 \right) \left( - \frac{1}{5} + 1 \right)^2 + 6\left( - \frac{1}{5} + 1 \right) \left( - \frac{1}{5} - 1 \right)^2 + 2 \left( - \frac{1}{5} - 1 \right)^3 - 6 \left( - \frac{1}{5} - 1 \right)^2 \left( - \frac{1}{5} + 1 \right)\]

\[ = \frac{576}{125} + \frac{384}{125} - \frac{432}{125} - \frac{864}{125} = \frac{- 336}{125} < 0\]

\[\text { So,} x = - \frac{1}{5} \text { is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( - \frac{1}{5} \right) = - \left( - \frac{1}{5} - 1 \right)^3 \left( - \frac{1}{5} + 1 \right)^2 = - \left( \frac{- 216}{125} \right)\left( \frac{16}{25} \right) = \frac{3465}{3125}\]

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Solution F(X) = − ( X − 1 ) 3 ( X + 1 ) 2 . Concept: Graph of Maxima and Minima.
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