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Model Paper 7 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 72 Marks
Question
Find the absolute maximum value and the absolute minimum value of the function
$f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$
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Answer


Given that  $f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$
$\Rightarrow f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x$
$\text { Now, } f(x)=0$
$\Rightarrow x=4$
Now, we evaluate the value of $f$ at critical point $x = 0$ and at end points of the interval
$f(4)=16-\frac{1}{2}(16)=16-8=8$
$f(-2)=-8-\frac{1}{2}(4)=-8-2=-10$
$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2$
$=18-\frac{81}{8}$
$=18-10.125$
$=7.875$
Therefore, the absolute maximum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is $8$ occurring at $x=4$
And, the absolute minimum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is $-10$ occurring at $x=-2$

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