PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# F(X) = 4x − X 2 2 in [ − 2,4,5] . - PUC Karnataka Science Class 12 - Mathematics

#### 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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Solution F(X) = 4x − X 2 2 in [ − 2,4,5] . Concept: Graph of Maxima and Minima.
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