Rolle’s Theorem: Statement, Questions & Answers, Interpretation

Rolle’s theorem is very useful in Calculus for continuous functions. In this post, we will learn Rolle’s theorem with its geometrical interpretation along with some solved examples.

Statement of Rolle’s Theorem

Let f(x) be a real-valued function defined on the closed interval [a, b] satisfying the following:

1. f(x) is continuous on [a, b]
2. f(x) is differentiable in (a, b), that is, f'(x) exists in (a, b).
3. f(a) = f(b).

Then there is a point c ∈ (a, b) such that f'(c) = 0.

Geometrical Interpretation

The geometric interpretation of Rolle’s theorem is the following: If f(x) is continuous on [a, b] and differentiable in (a, b) with f(a) = f(b), then according to the Rolle’s theorem there is a point c in the open interval (a, b) where the slope of the tangent to the curve y=f(x) is zero, that is, the tangent to the curve at the point (c, f(c)) is parallel to the x-axis.

Questions and Answers

Now we will learn how to apply Rolle’s theorem to various functions. A list of problems on Rolle’s theorem is given below with full solution.

Rolle’s Theorem for a Polynomial Function

$\boxed{\color{blue}\textbf{Question 1:}}$ Verify Rolle’s theorem for f(x) = x²-5x+6 in the interval 1 ≤ x ≤ 4.

$\boxed{\color{red}\textbf{Answer:}}$

Here f(x) = x²-5x+6, a=1, b=4.

As f(x) = x²-5x+6 is a polynomial function, it is continuous on [1, 4] and differentiable in (1, 4) with its derivative

f'(x) = 2x-5.

Also,

f(1) = 12 -5⋅1 +6 = 1-5+6 = 2. f(4) = 42 -5⋅4 +6 = 16-20+6 = 2.

Therefore, f(1) = f(4). So we conclude that all the three conditions of Rolle’s theorem are satisfied for f(x). Now,

f'(c) = 0

⇒ 2c-5 = 0

⇒ c = 5/2 ∈ (1, 4).

Thus, the Rolle’s theorem is verified for f(x) = x²-5x+6 in the interval 1 ≤ x ≤ 4.

Rolle’s Theorem for sin x

$\boxed{\color{blue}\textbf{Question 2:}}$ Verify Rolle’s theorem for f(x) = sinx on [0, π].

$\boxed{\color{red}\textbf{Answer:}}$

Here f(x) = sinx, a=0, b=π.

Note that f(x) is continuous on [0, π]. It is also differentiable in (0, π) with derivative f'(x) = cosx. Also, sin0 = sinπ = 0. Hence, f(x)=sinx satisfies all the conditions of Rolle’s theorem, and we can apply Rolle’s theorem. Now,

f'(c) = 0

⇒ cos c = 0

⇒ c = π/2 ∈ (0, π).

So the Rolle’s theorem is verified for f(x) = sin x in the interval 0 ≤ x ≤ π.

Rolle’s Theorem for tan x

$\boxed{\color{blue}\textbf{Question 3:}}$ Verify Rolle’s theorem for the function f(x) = tanx on [0, π].

$\boxed{\color{red}\textbf{Answer:}}$

Here f(x) = tanx, a=0, b=π.

Note $f(x) = \dfrac{\sin x}{\cos x}$.

Thus, f(x) won’t be continuous at the points where cosx =0. Observe that cosx=0 at the point x = π/2 in the interval (0, π).

So f(x) = tan x is not continuous at the point x = π/2. Hence, we cannot apply Rolle’s theorem in the interval 0 ≤ x ≤ π.

Rolle’s Theorem for cos x

$\boxed{\color{blue}\textbf{Question 4:}}$ Verify Rolle’s theorem for f(x) = cosx in the interval -π/2 ≤ x ≤ π/2.

$\boxed{\color{red}\textbf{Answer:}}$

Here f(x) = cosx, a=-π/2, b=π/2.

The function f(x) = cos x is continuous on [-π/2, π/2] and differentiable in (-π/2, π/2) with its derivative

f'(x) = -sinx.

Also, cos(-π/2) = cos(π/2) = 0 ⇒ f(-π/2) = f(π/2). So f(x)=cosx satisfies all the three conditions of Rolle’s theorem. Now,

f'(c) = 0 ⇒ -sin c = 0 ⇒ c = 0 and it lies in the interval (-π/2, π/2).

So the Rolle’s theorem is verified for f(x) = cos x in the interval -π/2 ≤ x ≤ π/2.

Rolle’s Theorem for Mod x

$\boxed{\color{blue}\textbf{Question 5:}}$ Verify Rolle’s theorem for f(x) = |x| in the interval -1 ≤ x ≤ 1.

$\boxed{\color{red}\textbf{Answer:}}$

The function f(x) =|x| is defined by $|x|=\begin{cases} x & \text{ if } x\geq 0 \\ -x & \text{ if } x<0. \end{cases}$

Here f(x) = |x|, a=-1, b=1.

Note f(x) = |x| is not differentiable at x=0 ∈ (-1, 1). This can be shown as follows:

The left-hand limit $L f'(0)$ = limx→0- $\dfrac{f(x)-f(0)}{x-0}$ = limx→0 $\dfrac{-x}{x}$ = -1. The right-hand limit $R f'(0)$ = limx→0+ $\dfrac{f(x)-f(0)}{x-0}$ = limx→0 $\dfrac{x}{x}$ = 1.

As the above two limits are different, we conclude that f(x)=|x| is not differentiable.

Thus, we cannot apply Rolle’s theorem to f(x) = |x| in the interval [-1, 1].

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