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Application of Derivatives — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives2 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 $f(x)=4 x-\frac{1}{2} x^{2}, x \in\left[-2, \frac{9}{2}\right]$ 
$\Rightarrow f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x$ 
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 $\left[-2, \frac{9}{2}\right]$ 
$f(4)=16-\frac{1}{2}(16)=16-8=8$ 
$f(-2)=-8-\frac{1}{2}(4)=-8-2=-10$
$\mathrm{f}\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}=18-10.125=7.875$ 
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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