3. Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

(viii) f (x) = x \sqrt{ 1-x } , 0 < x < 1

Answers (1)

Given function is
f (x) = x \sqrt{ 1-x }
f ^{'}(x) = \sqrt{1-x} + \frac{x(-1)}{2\sqrt{1-x}}
            = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} \Rightarrow \frac{2-3x}{2\sqrt{1-x}}\\ f^{'}(x) = 0\\ \frac{2-3x}{2\sqrt{1-x}} = 0\\ 3x = 2\\ x = \frac{2}{3}
Hence, x = \frac{2}{3} is the only critical point
Now,  we use the second derivative test
f^{''}(x)= \frac{(-1)(2\sqrt{1-x})-(2-x)(2.\frac{-1}{2\sqrt{1-x}}(-1))}{(2\sqrt{1-x})^2}
             = \frac{-2\sqrt{1-x}-\frac{2}{\sqrt{1-x}}+\frac{x}{\sqrt{1-x}}}{4(1-x)}
             = \frac{3x}{4(1-x)\sqrt{1-x}}
f^{"}(\frac{2}{3}) > 0
Hence, it is the point of minima and the minimum value is
f (x) = x \sqrt{ 1-x }\\ f(\frac{2}{3}) = \frac{2}{3}\sqrt{1-\frac{2}{3}}\\ f(\frac{2}{3}) = \frac{2}{3}\sqrt{\frac{1}{3}}\\ f(\frac{2}{3}) = \frac{2}{3\sqrt3}\\ f(\frac{2}{3}) = \frac{2\sqrt3}{9}

 

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