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Verify Rolle's theorem for f(x) = x3(x - 1)2 in [0,1]

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Given, f(x) = x3(x - 1)

f'(x) = 3x2(x - 1)2 + x2(x - 1) = x2(x - 1) (3(x - 1) + 2x)

or, f(x) = x2(x - 1)(5x - 3)

Clearly, f(x) is differentiable and continuous for all n.

So, (ii) f(x) is differentiable in (0,1) also 

(iii) f(0) = 0 and f(1) = 0

∴ f(0) = f(1)

Hence, all the condition of Rolle's theorem are satisfied for f(x) in [0,1].

So, f'(C) = 0

c2(c - 1)(5c - 3) = 0

c = 0, 1, 3/5

But 0 < c < 1

∴ c = 3/5

Thus, there exist at least one c, c = 3/5 between 0 & 1 such that f'(C) = 0.

Hence, Rolle's theorem has been verified

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