This is seventh 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:

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

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 $=f(0)$

$\qquad\qquad\qquad\quad\quad\;\;\;= \large\frac{1}{2}$

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