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)].

#### 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

#### 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

# 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)  #### 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,

$$f (x) = 0$$ 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'  #### 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

# 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'.

#### 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

#### 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