Mean Value Theorem Practice Problems

Mean Value Theorem (MVT) is a fundamental concept in calculus which is useful in both differential and integral calculus. Mean Value Theorem guarantees the existence of at least one point where the instantaneous rate of change (derivative) of a function equals the average rate of change over a given interval. Applying the Mean Value Theorem MVT through practice problems is crucial for mastering calculus and its applications in various engineering fields.

Various practice problems involving the Mean Value Theorem (MVT) are added below in the article, before starting with let's learn in brief about the Mean Value Theorem(MVT) first.

What is Mean Value Theorem?

Mean Value Theorem states that:

Let there be a function f(x) which satisfies the given condition:

• f(x) is continuous on the closed interval [a,b]
• f(x) is differentiable on the open interval (a,b).

Then there exist at least one number c in the interval (a,b) for which :

f′(c) = {f(b) - f(a)}/{(b - a)}

or

f′(c)(b - a) = f(b) - f(a)

This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem.

Mean Value Theorem Practice Problems with Solution

1. Verify Mean Value Theorem for the function f(x) = x^2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Given,

• f(x) = x² – 4x – 3

f'(x) = 2x – 4

a = 1 and b = 4 (given)

f(a) = f(1) = (1)2 – 4(1) – 3 = 1 – 4 – 3 = -6

f(b) = f(4) = (4)2 – 4(4) – 3 = -3

Now,

[f(b) – f(a)]/ (b – a) = (-3 + 6)/(4 – 1) = 3/3 = 1

As per the mean value theorem statement, there is a point c ∈ (1, 4) such that f'(c) = [f(b) – f(a)]/ (b – a), i.e. f'(c) = 1.

2c – 4 = 1

2c = 5

c = 5/2 ∈ (1, 4)

Verification: f'(c) = 2(5/2) – 4 = 5 – 4 = 1

Hence, verified the mean value theorem.

2. Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals: f (x) = x +1 / x, x ∈ [ −1, 2].

f(x)= x/1+x, x ∈ [ −1, 2]

f(0) = undefined

Therefore f(x) is not continuous at x = 0

Hence, Lagrange’s mean value theorem is not applicable.

3. Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals :f (x) = | 3x +1 |, x ∈ [ −1, 3].

f (x) = | 3x +1 |, x ∈ [ −1, 3]

3x + 1 = 0

x = -1/3

f(x) is not differentiable at x = -1/3

Hence, Lagrange’s mean value theorem is not applicable.

4. Verify Mean Value Theorem for the function f(x) = 4x^3 – 4x in the interval [a, b], where a = 0 and b = 3.

Given,

• f(x) = 4x³ – 4x

f'(x) = 12x² – 4

a = 0 and b = 3 (given)

f(a) = f(0) = 4(0)³ – 4(0) = 0 – 0 = 0

f(b) = f(3) = 4(3)³ – 4(3) = 108 – 12 = 96

Now,

[f(b) – f(a)]/ (b – a) = (96 - 0)/(3 – 0) = 96/3 = 32

As per, mean value theorem statement, there is a point c ∈ (0, 3) such that f'(c) = [f(b) – f(a)]/ (b – a), i.e. f'(c) = 32.

12c² – 4c = 32

12c² - 4c - 32 = 0

c = {1 + √(97)}/6 and c = {1 - √(97)}/6

c = {1 + √(97)}/6 ∈ (1, 4)

Hence, verified the mean value theorem.

5. Verify Mean Value Theorem for the function f(x) = x^3 – 3x + 2 in the interval [a, b], where a = 1 and b = 3.

Given,

• f(x) = x³ – 3x + 2

f'(x) = 3x² – 3

a = 1 and b = 3 (given)

f(a) = f(1) = (1)³ – 3.1 + 2 = 1 – 3 + 2 = 0

f(b) = f(3) = (3)³ – 3.3 + 2 = 27 – 9 + 2 = 20

Now,

[f(b) – f(a)]/ (b – a) = (20 - 0)/(3 – 1) = 20/2 = 10

As per the mean value theorem statement, there is a point c ∈ (1, 3) such that f'(c) = [f(b) – f(a)]/ (b – a), i.e. f'(c) = 10.

3c² – 3 = 10

3c² = 13

c² = 13/3

c = ±√(13/3)

c = √(13/3) ∈ (1, 4)

c = √(13/3) is within the interval (1, 3).

Hence, verified the mean value theorem.

Mean Value Theorem Worksheet

Q1. Using Mean Value Theorem find all possible values of c: f(x) = 3 + √x on [0, 4]

Q2. Using Mean Value Theorem find all possible values of c: f(x) = x^2(x - 1) on [0, 3]

Q3. Using Mean Value Theorem find all possible values of c: f(x) = x^2 - x^{2/3} on [-1, 8]

Q4. Using Mean Value Theorem find all possible values of c: f(x) = x/1 + x on [1, 3]

Q5. Using Mean Value Theorem find all possible values of c: f(x) = sin2x on [0, π]

Q6. Using Mean Value Theorem find all possible values of c: f(x) = x + 3cosx on [-π, π]