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Verify Rolle's theorem for the following functions; $f(x)=|x-1|,0\leq x\leq 2.$

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
answered Jul 24, 2013 by meena.p

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