# Verify Rolle’s theorem for the function f(x) = $$x^2$$ – 5x + 6 on the interval [2, 3].

## Solution :

Since a polynomial function is everywhere differentiable and so continuous also.

Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3).

Also, f(2) = $$2^2$$ – 5 $$\times$$ 2 + 6 = 0 and f(3) = $$3^2$$ – 5 $$\times$$ 3 + 6 = 0

$$\therefore$$ f(2) = f(3)

Thus, all the conditions of rolle’s theorem are satisfied. Now, we have to show that there exists some c $$\in$$ (2, 3) such that f'(c) = 0.

for this we proceed as follows,

We have,

f(x) = $$x^2$$ – 5x + 6 $$\implies$$ f'(x) = 2x – 5

$$\therefore$$ f'(x) = 0 $$\implies$$ 2x – 5 = 0 $$\implies$$ x = 2.5

Thus, c = 2.5 $$\in$$ (2, 3) such that f'(c) = 0. Hence, Rolle’s theorem is verified.

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