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# Find the Absolute Maximum and Minimum Values of a Function F Given by F ( X ) = 12 X 4 / 3 − 6 X 1 / 3 , X ∈ [ − 1 , 1 ] . - CBSE (Arts) Class 12 - Mathematics

#### Question

Find the absolute maximum and minimum values of a function f given by f(x) = 12 x^(4/3) - 6 x^(1/3) , x in [ - 1, 1] .

#### Solution

$\text { Given}: f\left( x \right) = 12 x^\frac{4}{3} - 6 x^\frac{1}{3}$

$\Rightarrow f'\left( x \right) = 16 x^\frac{1}{3} - 2 x^\frac{- 2}{3} = \frac{2\left( 8x - 1 \right)}{x^\frac{2}{3}}$

$\text { For a local maximum or a local minimum, we must have }$

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

$\Rightarrow \frac{2\left( 8x - 1 \right)}{x^\frac{2}{3}} = 0$

$\Rightarrow 8x - 1 = 0$

$\Rightarrow x = \frac{1}{8}$

$\text { Thus, the critical points of f are } - 1, \frac{1}{8} \text { and }1 .$

$\text { Now },$

$f\left( - 1 \right) = 12 \left( - 1 \right)^\frac{4}{3} - 6 \left( - 1 \right)^\frac{1}{3} = 18$

$f\left( \frac{1}{8} \right) = 12 \left( \frac{1}{8} \right)^\frac{4}{3} - 6 \left( \frac{1}{8} \right)^\frac{1}{3} = \frac{- 9}{4}$

$f\left( 1 \right) = 12 \left( 1 \right)^\frac{4}{3} - 6 \left( 1 \right)^\frac{1}{3} = 6$

$\text { Hence, the absolute maximum value when x = - 1 is 18 and the absolute minimum value when } x = \frac{1}{8}\text{ is }\frac{- 9}{4} .$

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Solution Find the Absolute Maximum and Minimum Values of a Function F Given by F ( X ) = 12 X 4 / 3 − 6 X 1 / 3 , X ∈ [ − 1 , 1 ] . Concept: Graph of Maxima and Minima.
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