Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

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- Rolle's Theorem: Let $f$ be a real valued function that satisfies the following conditions:
- (i) f is defined and continuous on the closed interval $[a,b]$
- (ii) f is differentiable in the open interval $(a,b)$
- (iii) $f(a)=f(b)$
- Then there exists at least one value $c \in (a,b)$ such that $f'(c)=0$

$f(x)=|x-1| \qquad 0\leq x \leq 2$

Step 1:

$f(x)= \left\{ \begin{array}{1 1} -(x-1) & \quad 0 \leq x \leq 1 \\ (x-1) & \quad 1 < x \leq 2 \end{array} \right. $

At $x=1 \qquad f(x)=0$.

The function is continuous in [0,2]

Step 2:

Consider $f'(x)\;at\; x=1$

The left derivative is -1

The right derivative is 1

The function is not differentiable at $x=0$

The conditions for Rolle's Theroem are not satisfied

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