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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: $ f (x) =\large\frac{x}{2}$$+\large\frac{2}{x}$$,x>0$

$\begin{array}{1 1} Maximum = 2 \\ Minimum= 2 \\ Minimum = 4 \\ Maximum= 0 \end{array} $

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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:
Given
$f (x) =\large\frac{x}{2}+\large\frac{2}{x}$
Differentiating with respect to x
$f'(x)=\large\frac{1}{2}-\frac{2}{x^2}$
$\qquad= \large\frac{x^2-4}{2x^2}$
$\qquad=\large\frac{(x-2)(x-2)}{2x^2}$
Now,$f'(x)=0\Rightarrow \large\frac{x^2-4}{2x^2}$$=0$
$x^2-4=0$
$x^2=4$
$x=\pm 2$
Step 2:
Since it is given that $x >0$ hence $x=-2$ is rejected.
At $x=2$
$x+2=2+2=4=+ve$
For $x$ slightly < 2
$f'(x)=\large\frac{(-)(+)}{(+)}$$=-ve$
For $x$ slightly > 2
$f'(x)=\large\frac{(+)(+)}{(+)}$$=+ve$
Step 3:
Thus $f'(x)$ changes sign from -ve to +ve.
Hence $f(x)$ has a local minima at $x=2$.
Local minimum value $f(x)=\large\frac{2}{2}+\frac{2}{2}$
$\Rightarrow 1+1$
$\Rightarrow 2$
answered Aug 7, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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