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Question
f(x) = 4x \[-\] \[\frac{x^2}{2}\] in [ \[-\] 2,4,5] .
Solution
\[\text { Given }: f\left( x \right) = 4x - \frac{x^2}{2}\]
\[ \Rightarrow f'\left( x \right) = 4 - x\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 4 - x = 0\]
\[ \Rightarrow x = 4\]
\[\text { Thus, the critical points of f are - 2, 4 and 4 . 5 } . \]
\[\text { Now }, \]
\[f\left( - 2 \right) = 4\left( - 2 \right) - \frac{\left( - 2 \right)^2}{2} = - 8 - 2 = - 10\]
\[f\left( 4 \right) = 4\left( 4 \right) - \frac{\left( 4 \right)^2}{2} = 16 - 8 = 8\]
\[f\left( 4 . 5 \right) = 4\left( 4 . 5 \right) - \frac{\left( 4 . 5 \right)^2}{2} = 18 - 10 . 125 = 7 . 875\]
\[\text { Hence, the absolute maximum value when x = 4 is 8 and the absolute minimum value when } x = - 2 \text{ is } - 10 . \]
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