Question

f(x) = 4x \[-\] \[\frac{x^2}{2}\] in [ \[-\] 2,4,5] .

Answer 1

\[\text { Given }: f\left( x \right) = 4x - \frac{x^2}{2}\]

\[ \Rightarrow f'\left( x \right) = 4 - x\]

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

\[f'\left( x \right) = 0\]

\[ \Rightarrow 4 - x = 0\]

\[ \Rightarrow x = 4\]

\[\text { Thus, the critical points of f are - 2, 4 and 4 . 5 } . \]

\[\text { Now }, \]

\[f\left( - 2 \right) = 4\left( - 2 \right) - \frac{\left( - 2 \right)^2}{2} = - 8 - 2 = - 10\]

\[f\left( 4 \right) = 4\left( 4 \right) - \frac{\left( 4 \right)^2}{2} = 16 - 8 = 8\]

\[f\left( 4 . 5 \right) = 4\left( 4 . 5 \right) - \frac{\left( 4 . 5 \right)^2}{2} = 18 - 10 . 125 = 7 . 875\]

\[\text { Hence, the absolute maximum value when x = 4 is 8 and the absolute minimum value when } x = - 2 \text{ is } - 10 . \]