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5. Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(iii) f (x) = 4 x - \frac{1}{2} x^2 , x \epsilon \left [ -2 , \frac{9}{2} \right ]

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Given function is 
f(x) =4x - \frac{1}{2}x^2
f^{'}(x) = 4 - x \\ f^{'}(x)= 0\\ 4-x= 0\\ x=4         
Hence, x = 4 is the critical point  of function f(x) =4x - \frac{1}{2}x^2
Now, we need to check the value of function f(x) =4x - \frac{1}{2}x^2  at  x = 4  and at the end points of given range  i.e. at x = -2 and x =  9/2
f(4) =4(4) - \frac{1}{2}(4)^2
            =16-\frac{1}{2}.16 = 16-8 = 8
f(-2) = 4(-2)-\frac{1}{2}.(-2)^2 = -8-2 = -10
f(\frac{9}{2}) =4(\frac{9}{2})-\frac{1}{2}.\left ( \frac{9}{2} \right )^2 = 18-\frac{81}{8} = \frac{63}{8}
Hence,  absolute maximum value of function f(x) =4x - \frac{1}{2}x^2  occurs at x = 4 and value is 8
and absolute minimum value of function f(x) =4x - \frac{1}{2}x^2  occurs at x = -2 and value is -10

Posted by

Gautam harsolia

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