Here you will learn statement of rolle’s theorem, it’s geometrical and algebraic interpretation with examples.

Let’s begin –

## Rolle’s Theorem

Statement: Let f be a function that satisfies the following three conditions:(a) f is continous on the closed interval [a, b].

(b) f is differentiable on the open interval (a, b)

(c) f(a) = f(b)

Then, there exist a real number c \(\in\) (a, b) such that f'(c) = 0.

**Geometrical Interpretation :**

Geometrically, the theorem says that somewhere between A and B the curve has atleast one tangent parallel to x-axis.

**Algebraic Interpretation :**

If f is differentiable function then between any two consecutive roots of f(x) = 0, there is atleast one root of the equation f'(x) = 0.

**Remark – **On this theorem generally two types of problems are formulated.

(a) To check the applicability of rolle’s theorem to a given function on a given interval.

(b) To verify rolle’s theorem for a given function on a given interval. In both types of problems we first check whether f(x) satisfies conditions of theorem or not. The following results are very helpful in doing so.

- A polynomial function is everywhere continuous and differentiable.
- The exponential function, sine and cosine functions are everywhere continuous and differentiable.
- Logarithmic function is continuous and differentiable in its domain.
- | x | is not differentiable at x = 0

**Example** : 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.