Rolle's Theorem: Statement, Geometrical Interpretation & Examples

Ans. y= x³ – 4x the function is continuous and differentiable in the given interval.

F(2)= f(-2)

F’(x)= 3x²- 4

F’(x)=0

3x²- 4=0

X= ± 2Ö3/ 3

Therefore, the two values of x satisfy Rolle’s Theorem.

Ans. f(x) = 2x (x - 3)n 

Differentiating the above mentioned function with respect to ‘x’

f’(x) = 2[xn (x - 3)n-1 + (x - 3)n ]

f’(x) = 2(x - 3)n [ xn/(x-3) + 1]

f’(c) = 2(c - 3)n [ cn/(c-3) + 1]

f’( ¾) = 0

2 - (9/4)n [ -n/3 + 1] = 0

-n/3 + 1 = 0

(-n + 3)/ 3 = 0

-n + 3 = 0

-n = -3

n = 3

Therefore, the required value of ‘n’ is 3.

Ans. The option a) is the correct answer

Explanation: given, f(x) = x³ - 3x

The above mentioned polynomial function is continuous and derivable in R.

Therefore, the function is continuous on [0, √3] and differentiable on [0, √3]

Differentiating the function with respect to x,

f(x) = x³ - 3x

f’(x) = 3x² - 3

Therefore, f’(c) = 3c² - 3

f’(c) = 0

3c² - 3 = 0

C² - 1 = 0

C² = 1

C = ± 1

Ans. The conditions for the applicability of Rolle’s theorem,

i) In the closed interval (a,b) the function f must be continuous.

ii) In the open interval (a,b) the function f must be differentiable

iii) If f (a) = f (b), then at least one value of x exists which lies between a and b i.e. a, and the value of c is calculated by f‘(c) = 0.

Now the given function is f(x) = sin 3x on (0, π)

⇒ f(0) = sin 3(0)

⇒ f(0) = sin(0)

⇒ f(0) = 0

⇒ f(π) = sin 3π

⇒ f(π) = sin (3π)

⇒ f(π) = 0

We got f(0) = f(π), so there exist a c π (0, π) such that f’(c) = 0

Now, the derivative of f(x)

⇒ f’(x) = d(sin 3x)/ dx

⇒ f’(x) = cos 3x d(3x)/ dx

⇒ f’(x) = 3 cos3x

We have f’(c) = 0

⇒ 3 cos3c = 0

⇒ 3c = π/2

⇒ c = π/6 π (0, π)

Thus, Rolle’s theorem is verified.

Ans. Given: f(x) = x² - 4x - 3 and a = 2, b = 3

f’(x) = 2x - 4

f(a) = f(2) = (2)² - 4(2) - 3 = 4 - 8 - 3 = -7

f(b) = f(3) = (3)² - 4(3) - 3 = -6 

Now, 

[f(b) - f(a)]/ (b-a) = (-6 + 7)/ (3 -2 ) = 1/1 = 1

According to the mean value theorem statement, there exists a point c ∈ (1, 4) such that f’(c) = [f(b) - f(a)]/ (b-a) or, f’(c) = 1

2c - 4 = 1

2c = 5

C = 5/2 ∈ (1, 4)

f’(c) = 2(5/2) - 4 = 5 - 4 = 1

Ans. Given: f(x) = x² - 4x - 3 and a = 1, b = 4

f’(x) = 2x - 4

f(a) = f(3) = (3)² - 4(3) - 3 = -6 

f(b) = f(5) = (5)² - 4(5) - 3 = 2 

Now, 

[f(b) - f(a)]/ (b-a) = (2 + 6)/ (5 - 3) = 8/2 = 4

According to the mean value theorem statement, there exists a point c ∈ (1, 4) such that f’(c) = [f(b) - f(a)]/ (b-a) or, f’(c) = 4

2c - 4 = 4

2c = 8

C = 8/2 ∈ (1, 4)

f’(c) = 2(8/2) - 4 = 8 - 4 = 4

Ans. Given, a = -2 and b = 3

• f(x) = 2x^{2 +} 1
• f'(x) = 2x
• f(a) = f(0) = 2(-2)² + 1
•  9
• f(b) = f(2) = 2(3)² + 1
• 19
• Now,
• f(b) = f(a) thus the condition for Rolle’s Theorem is verified.
• we know that, f'(c) = 0
• 4c + 1 = 0
• 4c = -1
• c = -1/4
• c = -1/4 ∈ (-2, 3)
• Hence, Rolle’s Theorem is verified.

Ans. According to Rolle’s theorem, the function y = x² + 2 is continuous and differentiable within (–2,2).

The given function,

F(x) = x² + 2

Putting the value of -1 we get,

F(−2) = −1² + 2 = 1 + 2 = 3

Putting the value of 3 we get,

F(2) = 1² + 2 = 1 + 2 = 3

Thus, f(–2) = f(2) = 3

So, from the equation, it is clear that the value of f(x) at –2 and 2 points coincide with each other.

Now, f'(x) = 2x

According to Rolle’s theorem, a point c ∈ –1,1 is defined in such a way that f′(c) = 0.

At point c = 0, f′(c) = 0, here c = 0 exists between –1,1. So, the theorem is verified.

Ans. F(x) = x^2/3

f ’(x) = 2/3x^1/3

f ’ (0) = 2 /3(0)^1/3

f ’(0) = ∞

So f ‘(x) does not exist at x=o and is not differentiable between (-2, 1)

So Rolle’s Theorem is not applicable on F(x) in (−2,1).

Ans. f (x) = x²-4x-3, the given variable is continuous in the interval (1,6) and derivable (1,6).

The polynomial satisfies all the conditions of Rolle’s Theorem.

The f ’(x) = 2x-4

f ’(x) = 2c-4

f (6) = 6²-4x6-3 = 9

f (1) = 1-4-3 = -6

Now,

f ’(x) = 2c-4

2c-4 = 0

C= 2

As, 2€ (1, 6) Rolle’s Theorem is applicable.

Ans. Rolle’s Theorem is satisfied if it satisfied the three conditions:

• f (x) = x²+2x-8 must be continuous at (2,2)
• f (x) = x²+2x-8 must be differentiable at (2,2)
• f(-2)= f(2)

f (x) = x²+2x-8 is a polynomial and continuous for all real values of x. So, f (x) = x²+2x-8 is continuous at (2,2).

For all real values of x, f (x) = x²+2x-8 must be differentiable at (2,2).

Now,

f(2)= (2)² +2(2) -8

=0

f (2)= (2)²+2(2) -8

=0

Hence,

f(-2)= f (2)

Now,

f (x) = x²+2x-8

f ‘(x) = 2x+2

f ‘(c) = 2c+2

Though all three conditions are satisfied, then

f ‘(c)=0

2c+2=0

c = -1

The value of c falls between (2,2).

Therefore Rolle’s Theorem is satisfied between the given limit.

Ans. The function y = x² + 1, as it is a polynomial function, is continuous in [– 3, 3] and differentiable in (–3, 3). Also,

f(-3) = (-3)² + 1 = 9 + 1 = 10

f(3) = (3)² + 1 = 9 + 1 = 10

Thus, f(– 1) = f(1) = 10

Hence, the function f(x) satisfies all conditions of Rolle's theorem.

Now, f'(x) = 2x

Rolle’s theorem states that there is a point c ∈ (– 3, 3) such that f′(c) = 0.

2c = 0

c = 0, where c = 0 ∈ (–3, 3)