Question
Verify Rolle's Theorem for the following function:

$f (x) = (x - 1) (x - 2)^{2}, [1,2]$

Answer

$f (x) = (x - 1) (x - 2)^{2}, [1,2]$

The function f(x) is differentiable on [1, 2] and so it is continuous on [1, 2]

$\therefore$ Al conditions of Rolles Theorem are satisfied

$\therefore f' (x) = (x - 1) 2 (x -2) + (x - 2)^{2}$

$= (x - 2)[2x - 2 + x- 2] = (x - 2) (3x - 4)$

$\therefore f' (c) = 0 \Rightarrow c = 2 \text{ and}\ {c}\frac{4}{3}$

$As 1<\frac{4}{3}<2, \text{the Roll's theorem is verfied}$

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