Informative line

Rolles Theorem

Learn Rolle’s theorem example in calculus, practice problems for finding the value of number 'c' which satisfies conclusion of Rolle's Theorem for a function.

Statement of Rolle's Theorem

Rolle's Theorem :

Let \('f'\) be a function defined in [a, b] such that

  1. It is continuous in closed interval [a, b]
  2. It is differentiable in open interval (a, b)
  3. \(f(a)=f(b)\), then there is at least one number 'c' in (a, b) such that f ' (c) = 0

  • Observe these two graphs fig(1) & fig (2). In figure (1) , there are three points at which the graph has a horizontal tangent (c1, c2, c3). In figure (2), there is only one such 'c'.
  • Rolle's theorem says that there is at least one 'c'. Number of c's will depend on the nature of the function.
  • Rolle's theorem says that if we join two points of equal height on a graph, there must be a turning point. [If the graph is not broken (continuous) or does not have a sharp corner (differentiable)].

Illustration Questions

Which one of the following functions doesn't satisfy the hypotheses of Rolle's theorem in the indicated interval?

A \(f(x)=2x^3+7x^2-9x\)  in [0, 1]

B \(f(x)=sinx\)  in \([0, \pi]\)

C \(f(x)=\dfrac {x-1}{x-4}\)  in [2, 5]

D \(f(x)=x^2+5x+6\)  in [–3, –2]

×

For option (A),

\(f(x)=2x^3+7x^2-9x\)  in [0, 1]

Since \('f'\)is a polynomial function, it is continuous and differentiable for all values of x.

  1. Continuous in [0, 1]
  2. Differentiable in (0, 1)
  3. \(f(0)=0\) and \(f(1)=2+7-9=0\)

       \(\Rightarrow f(1)=f(0)\)

\(\therefore\) Rolle's theorem is applicable.

For option (B),

\(f(x)=sinx\)  in \([0, \pi]\)

Since \('f'\)is continuous and differentiable for all real values of x.

  1. Continuous in \([0, \pi]\)
  2. Differentiable in \((0, \pi)\)
  3. \(f(0)=sin0=0\) and \(f(\pi)=sin\pi=0\)

       \(\Rightarrow f(0)=f(\pi)=0\)

\(\therefore\) Rolle's theorem is applicable.

For option (C),

  \(f(x)=\dfrac {x-1}{x-4}\)  in [2, 5]

Observe that \('f'\)has a point of discontinuity at x = 4.

\(\therefore f\) is discontinuous at  [2, 5]

\(\therefore \) Rolle's theorem is not applicable in [2, 5]

For option (D),

\(f(x)=x^2+5x+6\)  in [–3, –2]

Since \('f'\)is  a polynomial function and it is continuous and differentiable for all real values of x.

  1. \('f'\) is continuous in [–3, –2]    
  2. \('f'\)is differentiable in (–3, –2)
  3. \(f(-3)=9-15+6=0\)
  4. \( f(-2)=4-10+6=0\)

           \(\Rightarrow f(-3)=f(-2)\)

\(\therefore\) Rolle's theorem is applicable.

Which one of the following functions doesn't satisfy the hypotheses of Rolle's theorem in the indicated interval?

A

\(f(x)=2x^3+7x^2-9x\)  in [0, 1]

.

B

\(f(x)=sinx\)  in \([0, \pi]\)

C

\(f(x)=\dfrac {x-1}{x-4}\)  in [2, 5]

D

\(f(x)=x^2+5x+6\)  in [–3, –2]

Option C is Correct

Finding the number c for a Function at which it satisfies Rolle's Theorem

  • Given a function \(f(x)\) defined in [a, b] which satisfies the hypotheses of Rolle's theorem then solve.,
  • \(f'(x)=0\; \text{in } [a, b]\),

  • The solutions to this equation (there can be more than one) will be called the number c (or c's)
  1. Find \(f'(x)\) i.e. derivative of function.
  2. Put \(f'(x)=0\).
  3. Find roots, these will be the values of 'c'.

Illustration Questions

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=x^2-4x+3\) in [1, 3]

A c= 3

B c=\(\dfrac {3}{2}\)

C c= 2

D c= 5

×

 \(f(x)=x^2-4x+3\) in [1, 3]

To find 'c', solve \(f'(x)=0\)

\(\Rightarrow 2x-4=0\;\Rightarrow x=2\)

 \(Since, \,2\in(1, 3)\; \Rightarrow c=2\)

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=x^2-4x+3\) in [1, 3]

A

c= 3

.

B

c=\(\dfrac {3}{2}\)

C

c= 2

D

c= 5

Option C is Correct

Illustration Questions

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=(x-1)(x-2)^2\) in [1, 2]

A c= 2

B c= 1.7

C c= \(\dfrac {4}{3}\)

D c= –5

×

 \(f(x)=(x-1)(x-2)^2\) in [1, 2]

  1. Find \(f'(x)\) i.e. derivative of function.
  2. Put \(f'(x)=0\)
  3. Find roots, these will be the values of 'c'.

To find 'c', solve \(f'(x)=0\)

\(\Rightarrow (x-1)×2(x-2)+1×(x-2)^2=0\)

\(\Rightarrow (x-2)[2(x-1)+(x-2)]=0\)

\(\Rightarrow (x-2)[3x-4]=0\)

\(\Rightarrow x=2, \;\dfrac {4}{3}\)

Since , \(2\notin(1, 2)\) and \(\dfrac {4}{3}\in(1, 2)\;\)

\(\Rightarrow\;c=\dfrac {4}{3}\)

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=(x-1)(x-2)^2\) in [1, 2]

A

c= 2

.

B

c= 1.7

C

c= \(\dfrac {4}{3}\)

D

c= –5

Option C is Correct

Illustration Questions

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=sinx+cosx\) in \(\left [0,\dfrac {\pi}{2}\right]\)

A c= \(\dfrac {\pi}{4}\)

B c= \(-\dfrac {\pi}{8}\)

C c= \(\dfrac {\pi}{6}\)

D c= \(\dfrac {\pi}{3}\)

×

\(f(x)=sinx+cosx\) in \(\left [0,\dfrac {\pi}{2}\right]\)

To find 'c', solve \(f'(x)=0\)

\(\Rightarrow cosx-sinx=0\;\\\Rightarrow sinx=cosx\)

\(\Rightarrow tanx=1\;\\\Rightarrow x = \dfrac {\pi}{4}\)

 

Since ,\(\dfrac {\pi}{4}\in\left (0,\;\dfrac {\pi}{2}\right)\)

\(\Rightarrow c=\dfrac {\pi}{4}\)

Find the number 'c' that satisfies the conclusion of Rolle's theorem for the function \(f(x)=sinx+cosx\) in \(\left [0,\dfrac {\pi}{2}\right]\)

A

c= \(\dfrac {\pi}{4}\)

.

B

c= \(-\dfrac {\pi}{8}\)

C

c= \(\dfrac {\pi}{6}\)

D

c= \(\dfrac {\pi}{3}\)

Option A is Correct

Problems on Number of Roots of an Equation

  • Consider the equation \(f(x)=0\) for some function \('f'\) whose derivative is always positive or negative i.e. \(f ' (x) \neq 0\)  for any value of \(x\),

          then \(f(a) = f(b)\) is not possible for distinct values of 'a' and 'b' as it will violate Rolle's theorem.

  • There cannot be two roots, there can be at most one root.

 

  • Further if p, q to be two values of x such that \(f(p)\) and  \(f(q)\) are of opposite signs then 

             by intermediate value theorem,

\(f(x)=0\), will have only one root and it will lie in (p, q)

Illustration Questions

The equation  \(5x+ sinx = 0\)  has

A exactly one real root

B exactly two real roots

C at least two real roots

D at least four real roots

×

The equation is \(5x + sinx = 0\)

Consider, \(f(x) = 5x + sinx\)

\(f'(x) = 5 + cosx\)

Now, \(f'(x) = 5 + cosx\) cannot be 0 for any value of x as \(cosx\in[-1, 1]\)

\(\therefore\) \(f '(x) \neq 0\) for any x . 

Let \((\alpha)\) and \((\beta)\) be the two roots of \(f(x) = 0\)

\(\therefore\)\(f '(x) = 0\) for at least one c in \((\alpha, \beta) \)  but this is not true,

therefore contradiction

\(f (x) = 0\) cannot have two roots. ...(i)

Now, \(f (-\pi) < 0\) and \(f(\pi)>0\)

\(\therefore\) By Intermediate Value Theorem

 \(f (x) = 0\)  has at least one root is  \((-\pi, \pi)\).

(Apply (i) )

\(\therefore\) \(f (x) = 0\) has exactly one root in \((-\pi, \pi)\)

 

The equation  \(5x+ sinx = 0\)  has

A

exactly one real root

.

B

exactly two real roots

C

at least two real roots

D

at least four real roots

Option A is Correct

Conditions at which Function does not Satisfy Rolle's Theorem

Function which does not satisfy Rolle's theorem will satisfy one of the following conditions 

(1) Discontinuous in [a, b]

(2) Non differentiable in (a, b)

(3) \(f(a)\neq f(b)\)

  • If f is discontinuous in [a, b] then there might not be any value of 'c' which satisfies  \(f' (c) = 0\)

  • Function which is discontinuous at \(x=\alpha\) and \(x=b\), there is no value of 'c'.
  • If \( f \) is continuous in [a, b) but non differentiable in (a, b), there might not be any value of c which satisfies  \( f'(c) = 0\).
  • \( f \) is non differentiable at \(x=\alpha\), there is no value of 'c'

Illustration Questions

Consider  \( f(x) = x^{2/3}\) in [–1, 1]. We see that  \( f(1) = f(-1) = 1\), but there is no \( 'c'\in(-1, 1)\) such that \( f'(c) = 0\). Why does this not contradict Rolle's theorem?

A \(f\) is discontinuous at \(x=0\)

B \(f\) is non differentiable at \(x=0\)

C \(f\) is discontinuous at \(x=\dfrac {1}{2}\)

D \(f\) is discontinuous at \(x=\dfrac {-1}{2}\)

×

For Rolle's theorem, there are three requirements

(1) \(f\) should be continuous in [a, b]

(2) \(f\) should be differentiable in (a, b)

(3) \( f(a) = f(b)\)

Now,

\(f(-1)=f(1)\) is true but \(f \)has non differentiable point at \(x = 0\)

\(\therefore\)  It is non differentiable in (–1, 1), so although (1) and (3) conditions are satisfied, (2) is violated 

\(\therefore\) Rolle's theorem is not applicable

Consider  \( f(x) = x^{2/3}\) in [–1, 1]. We see that  \( f(1) = f(-1) = 1\), but there is no \( 'c'\in(-1, 1)\) such that \( f'(c) = 0\). Why does this not contradict Rolle's theorem?

A

\(f\) is discontinuous at \(x=0\)

.

B

\(f\) is non differentiable at \(x=0\)

C

\(f\) is discontinuous at \(x=\dfrac {1}{2}\)

D

\(f\) is discontinuous at \(x=\dfrac {-1}{2}\)

Option B is Correct

Practice Now