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# 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}$$,x>0 This is eighth part of multipart q3 Can you answer this question? ## 1 Answer 0 votes Toolbox: • f is said to have a maximum value in I , if there exist a point c in I such that f(c) \geq f (x) for all x \in I.The number f( c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I • f is said to have a minimum value in I , if there exist a point c in I such that f(c) \leq f (x) for all x \in I.The number f(c) is called the minimum value of f in I and the point c in this case is called a point of minimum value of f in I • f is said to have a extreme value in I , if there exist a point c in I such that f(c) is either a maximum value or minimum value of f in I. The number f (c) in this case is called the extreme value of f in I and the point c is called the extreme point. Step 1: Let f(x)=x\sqrt{1-x},$$x >0$
Differentiating with respect to x
Apply product rule: d/dx(uv) = uv'+vu'
$f'(x)=1.\sqrt{1-x}+\large\frac{1}{2\sqrt{1-x}}$
On simplifying we get,
$\qquad=\large\frac{2(1-x)+(-x)}{2\sqrt{1-x}}$
$\qquad=\large\frac{2-2x-x}{2\sqrt{1-x}}$
$\qquad=\large\frac{2-3x}{2\sqrt{1-x}}$
$\qquad=\large\frac{-3(x-2/3)}{2\sqrt{1-x}}$
Step 2:
Now,$f'(x)=0$
$\qquad\qquad=\large\frac{-3(0-\Large\frac{2}{3})}{2\sqrt{1-0}}$
$x=\large\frac{2}{3}$
At $x=\large\frac{2}{3}$
$1-x=1-\large\frac{2}{3}$
$\qquad=\large\frac{1}{3}$$=+ve When x is slightly < \large\frac{2}{3} f'(x)=\large\frac{(-)(-)}{(+)}$$=+ve$
When $x$ is slightly > $\large\frac{2}{3}$
$f'(x)=\large\frac{(-)(+)}{(+)}$$=-ve Step 3: \therefore f'(x) changes sign from +ve to -ve as x increases through x=\large\frac{2}{3} \therefore f(x) has a local maxima at x=\large\frac{2}{3} Local maximum value =f\big(\large\frac{2}{3}\big) \qquad\qquad\qquad\qquad=\large\frac{2}{3}$$\sqrt{1-\large\frac{2}{3}}$
$\qquad\qquad\qquad\qquad=\large\frac{2}{3}\frac{1}{\sqrt 3}$
$\qquad\qquad\qquad\qquad=\large\frac{2\sqrt 3}{9}$
answered Aug 7, 2013
edited Aug 30, 2013

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