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

#### Question

f(x) = (x+1) (x+2)^(1/3), x>=-2 .

#### Solution

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

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

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

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

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

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

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

$\Rightarrow x + 1 = - 3x - 6$

$\Rightarrow x = \frac{- 7}{4}$

Thus, x = (- 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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