## 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.

### Similar Questions

Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\).