logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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: $ (vii)\: f (x) = \large\frac{1}{x^2+2}$

This is seventh 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:
$f(x)=\large\frac{1}{x^2+2}$
Differentiating with respect to x we get
$f'(x)=\large\frac{2x}{(x^2+2)^2}$
For Maxima and minima
Now $f'(x)=0$
$\large\frac{2x}{(x^2+2)^2}$$=0$
$\Rightarrow x=0$
Step 2:
At $x=0$
When $x$ is slightly < 0
$f'(x)=\large\frac{(-)(-)}{(+)}$
$\qquad=+ve$
When $x$ is slightly > 0
$f'(x)=\large\frac{(-)(+)}{(+)}$
$\qquad=-ve$
Step 3:
$\therefore f'(x)$ changes from +ve to -ve as $x$ increases through 0.
$\Rightarrow f(x)$ has a local maximum at $x=0$.
$\therefore$ Local maximum value $=f(0)$
$\qquad\qquad\qquad\quad\quad\;\;\;= \large\frac{1}{2}$
answered Aug 7, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...