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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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

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  • $f$ 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 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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