In this post, we will learn about Rolle’s theorem along with examples. We will also discuss some of the related questions of Rolle’s theorem with answers.

## Rolle’s Theorem Statement

Let f(x) be a real-valued function defined on [a, b]. If

- f(x) is continuous on [a, b]
- f(x) is differentiable in (a, b)
- f(a)=f(b),

then there exists at least one element c in (a, b) such that

$f'(c)=0$.

## Examples of Rolle’s Theorem

Let f(x)= $x\sqrt{1-x^2}$ in [0, 1]. See that

- f(x) is continuous on [0, 1]
- $f'(x)= \dfrac{1-2x^2}{\sqrt{1-x^2}}$ exists in (0, 1)
- f(0)=0=f(1).

So all the conditions of Rolle’s are satisfied. According to this theorem, we have c in (0, 1) such that

$f'(c)=0$.

Here c=1/√2. It lies between 0 and 1.

## Questions and Answers

**Question 1:** Check the applicability of Rolle’s theorem where f(x)=|x| in [-1, 1].

**Answer:**

See that 0 ∈ (-1, 1). So f(x)=|x| has to be differentiable at x=0 in order to apply Rolle’s theorem.

Lf'(0) = lim_{x→0-} $\dfrac{f(x)-f(0)}{x}$ = lim_{x→0} -x/x = -1

Rf'(0) = lim_{x→0+} $\dfrac{f(x)-f(0)}{x}$ = lim_{x→0} x/x =1

So f'(0) does not exist. Hence, we cannot apply Rolle’s theorem for f(x)=|x| in [-1, 1].

**Question 2:** Prove that between any two real roots of e^{x} sinx =1, there is at least one real root of e^{x} cosx = -1

**Answer:**

Let f(x) = e^{-x} – sinx.

Let α, β be two real roots of e^{x} sinx -1=0. That is,

e^{α} sinα -1=0 and e^{β} sinβ -1=0.

⇒ e^{α} sinα =1 and e^{β} sinβ =1

⇒ e^{-α} = sinα and e^{-β} = sinβ

Therefore, f(α)= f(β) =0.

We assume that α<β.

Here, f'(x) = -e^{-x} – cosx. Thus, we obtain that

- f(x) is continuous on [α, β]
- f(x) is derivable in (α, β)
- f(α)= f(β)

So by Rolle’s theorem, there is at least one c ∈ (α, β) such that

$f'(c)=0$

⇒ -e^{-c} – cosc = 0

⇒ -e^{-c} = cosc

⇒ e^{c} cosc = -1.

In other words, c is a root of e^{x}cosx = -1. Thus, we have shown that between any two real roots of e^{x} sinx =1, there is at least one real root of e^{x} cosx = -1.