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Solution for F(X) = (X+1) (X+2)1/3, X > − 2 . - CBSE (Science) Class 12 - Mathematics

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Question

f(x) = (x+1) (x+2)1/3, \[x\frac{>}{} - 2\] .

Solution

\[\text{Given:} \hspace{0.167em} f\left( x \right) = \left( x + 1 \right) \left( x + 2 \right)^\frac{1}{3} \]

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

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

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

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

\[ \Rightarrow \frac{1}{3}\left( x + 1 \right) = - \left( x + 2 \right)^\frac{1}{3} \times \left( x + 2 \right)^\frac{2}{3} \]

\[ \Rightarrow \frac{1}{3}\left( x + 1 \right) = - \left( x + 2 \right)\]

\[ \Rightarrow x + 1 = - 3x - 6\]

\[ \Rightarrow x = \frac{- 7}{4}\]

\[\text { Thus, x = \frac{- 7}{4} is the possible point of local maxima or local minima }. \]

\[\text { Now,} \]

\[f''\left( \frac{- 7}{4} \right) = \frac{2}{3} \left( x + 2 \right)^\frac{- 2}{3} - \frac{2}{9}\left( x + 1 \right) \left( x + 2 \right)^\frac{- 5}{3} \]

\[\text { At } x = \frac{- 7}{4}: \]

\[ f''\left( \frac{- 7}{4} \right) = \frac{2}{3} \left( \frac{- 7}{4} + 2 \right)^\frac{- 2}{3} - \frac{2}{9}\left( \frac{- 7}{4} + 1 \right) \left( \frac{- 7}{4} + 2 \right)^\frac{- 5}{3} = \frac{2}{3} \left( \frac{1}{4} \right)^\frac{- 2}{3} + \frac{1}{18} \left( \frac{1}{4} \right)^\frac{- 5}{2} > 0\]

\[\text { So}, x = \frac{- 7}{4} \text { is the point of local minimum }. \]

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

\[f\left( \frac{- 7}{4} \right) = \left( \frac{- 7}{4} + 1 \right) \left( \frac{- 7}{4} + 2 \right)^\frac{1}{3} = \frac{- 3}{4} \left( \frac{1}{4} \right)^\frac{1}{3} = \frac{- 3}{4^\frac{4}{3}}\]

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Solution for question: F(X) = (X+1) (X+2)1/3, X > − 2 . concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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