Note: This is part 4th of a 5 part question, split as 5 separate questions here.

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- (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
- (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$

Step 1:

$f(x)=x(x-1)(x+1)$ on $[-2,-1]$

$\qquad=x(x^2-1))$

$\qquad=x^3-x$

Step 2:

$f'(x)=3x^2-1$

$f'(x)=3 \bigg(x^2 -\large\frac{1}{3}\bigg)$

$f'(x)=3 \bigg(x -\large\frac{1}{\sqrt 3}\bigg) \bigg(x +\large\frac{1}{\sqrt 3}\bigg)$

$f'(x) >0$ when x lies outside $ \bigg (-\large\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\bigg)$

Step 3:

$[-2,-1]$ does not have any values in common with $ \bigg (-\large\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\bigg)$

$\therefore f'(x) >0$ for all $ x \in [-2,-1]$

$\therefore \;f$ is strictly increasing on $[-2,-1]$

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