Rolle’s Theorem Questions and Answers

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

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

1. f(x) is continuous on [0, 1]
2. $f'(x)= \dfrac{1-2x^2}{\sqrt{1-x^2}}$ exists in (0, 1)
3. 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

1. f(x) is continuous on [α, β]
2. f(x) is derivable in (α, β)
3. 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^xcosx = -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.