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) = 22 – 5 × 2 + 6 = 0 and f(3) = 32 – 5 × 3 + 6 = 0
∴ f(2) = f(3)
Thus, all the conditions of rolle’s theorem are satisfied. Now, we have to show that there exists some c ∈ (2, 3) such that f'(c) = 0.
for this we proceed as follows,
We have,
f(x) = x2 – 5x + 6 ⟹ f'(x) = 2x – 5
∴ f'(x) = 0 ⟹ 2x – 5 = 0 ⟹ x = 2.5
Thus, c = 2.5 ∈ (2, 3) such that f'(c) = 0. Hence, Rolle’s theorem is verified.
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