This is second 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:

$g(x)=x^3-3x$

Let $g(x)=x^3-3x$

Differentiating with respect to $x$

$g'(x)=3x^2-3$

$\quad\;\;\;\;=3[x^2-1]$

$\quad\;\;\;\;=3(x+1)(x-1)$

Now $g'(x)=0$

$\Rightarrow 3(x-1)(x+1)=0$

$\Rightarrow$ either $x=1$ or $x=-1$

Step 2:

At $\;x=1$

When $x$ is slightly <1

$g'(x)$ is -ve

When $x$ is slightly >1

$g'(x)$ is +ve

Thus $g'(x)$ changes sign from -ve to +ve as $x$ increases through 1 and hence $x=1$ is a point of local minimum.

Step 3:

At $\;x=-1$

When $x$ is slightly <-1

$g'(x)$ is +ve

When $x$ is slightly >-1

$g'(x)$ is -ve

Thus $g'(x)$ changes sign from positive to negative as $x$ increases through -1 and hence $x=-1$ is a point of local maximum.

Local minimum value =$f(1)=1-3=-2$

Local maximum value =$f(-1)=-1+3=2$

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