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

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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