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Verify Rolle’s theorem for each of the following functions:

Show that f(x) = x(x - 5)2 satisfies Rolle’s theorem on [0, 5] and that the value of c is (5/3)

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Condition (1):

Since, f(x) = x(x - 5)2 is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x(x - 5)2 is continuous on [0,5].

Condition (2):

Here, f’(x) = (x-5)2+ 2x(x - 5) which exist in [0,5].

So, f(x) = x(x - 5)2 is differentiable on (0,5).

Condition (3):

Here, f(0) = 0(0 - 5)= 0

And f(5) = 5(5 - 5)= 0

i.e. f(0) = f(5)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,5) such that f’(c) = 0

i.e. (c - 5)2+ 2c(c - 5) = 0

i.e.(c - 5)(3c - 5) = 0

i.e. c = 5/3 or c = 5

Value of c = 5/3 ∈ (0, 5)

Thus, Rolle’s theorem is satisfied.

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