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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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