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

#### 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 F(X) = (X+1) (X+2)1/3, X > − 2 . Concept: Graph of Maxima and Minima.
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