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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Find the absolute maximum value and the absolute minimum value of the following function in the given intervals:
$\text{f}\text{(x)}=4\text{x}-\frac{1}{2}\text{x}^2 ,\ \text{x}\in\Big[-2,\frac{9}{2}\Big]$
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Answer

Given: $\text{f}\text{(x)}=4\text{x}-\frac{1}{2}\text{x}^2,\ \text{x}\in\bigg(-2,\frac{9}{2}\bigg)$ $\Rightarrow\ \text{f}'\text{(x)}=4-\frac{1}{2} (2x)=4-\text{x}$
$\text{Now }\text{f}'\text{(x)}=0\ \Rightarrow\ 4-\text{x}=0$ $\Rightarrow\ \text{x}=4\in\bigg(-2,\frac{9}{2}\bigg)$
$\text{At }\text{ x}=4,\ \ \text{f}(4)=16-\frac{1}{2}(16)=16-8=8$
$\text{At }\text{ x}=-2,\ \ \text{f}(-2)=4(-2)-\frac{1}{2}(4)=-8-2=-10$
$\text{At }\text{ x}=\frac{9}{2},\ \text{f}\bigg(\frac{9}{2}\bigg)=4\bigg(\frac{9}{2}\bigg)-\frac{1}{2}\bigg(\frac{9}{2}\bigg)^2=18-\frac{81}{8}=\frac{63}{8}$
Therefore, absolute minimum value is -10 and absolute maximum value is 8.

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