Rolle’s Theorem – Statement and Examples

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.

  1. A polynomial function is everywhere continuous and differentiable.
  2. The exponential function, sine and cosine functions are everywhere continuous and differentiable.
  3. Logarithmic function is continuous and differentiable in its domain.
  4. | 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.


Related Questions

It is given that for the function f(x) = \(x^3 – 6x^2 + ax + b\) on [1, 3], Rolles’s theorem holds with c = \(2 +{1\over \sqrt{3}}\). Find the values of a and b, if f(1) = f(3) = 0.

Find the point on the curve y = cos x – 1, x \(\in\) \([{\pi\over 2}, {3\pi\over 2}]\) at which tangent is parallel to the x-axis.

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