Mean Value Theorem

The Mean Value Theorem states that for a curve passing through two given points there exists at least one point on the curve where the tangent is parallel to the secant passing through the two given points. Mean Value Theorem is abbreviated as MVT.

This theorem was first proposed by an Indian Mathematician Parmeshwara early 14th century. After this various mathematicians from all around the world worked on this theorem and the final theorem was proposed by Augustin Louis Cauchy in the year 1823.

In this article, we will learn about the Mean Value Theorem, its proof, its geometrical interpretation, and others in detail in this article.

What is the Mean Value Theorem?

The Mean Value Theorem states that for any function f(x) passing through two given points [a, f(a)], [b, f(b)], there exists at least one point [c, f(c)] on the curve such that the tangent through that point is parallel to the secant passing through the other two points.

In calculus, for a function f(x) defined on [a, b] → R, such that it is continuous and differentiable across an interval. Then mean value theorem is applicable when

• Function f(x) is continuous over the interval [a, b].
• Function f(x) is differentiable over the interval (a, b).

Then, there exists a point c in the interval (a, b) such that 

f'(c) = [f(b) - f(a)] / (b - a)

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Mean Value Theorem Proof

Statement: If a function f(x) is continuous over the closed interval [a, b], and differentiable over the open interval (a, b), then there exists at least one point c in the interval (a, b) such that f '(c) is zero, i.e. the tangent to the curve at point [c, f(c)] is parallel to the x-axis.

Proof: 

We know that the secant line to f(x) passes through [a, f(a)] and [b, f(b)]. 

Now, the equation of the line passing through two given points (x₁, y₁) and (x₂, y₂) is y - y₁ = [(y₂ - y₁)/(x₂ - x₁)](x - x₁)

The equation of the secant line is,

F(x) - f(a) = [f(b) - f(a)] / (b - a) (x-a)

F(x) = [ f(b) - f(a) ] / (b - a) (x-a) + f(a) ...(i)

Let, h(x) be F(x) - g(x)

h(x) = F(x) - [[ f(b) - f(a) ] / (b - a) (x-a) + f(a)]                    (From (i))

As, h(x) is continuous on [a, b] and differentiable on (a, b) and h(a) = h(b) = 0

Applying, Rolle's theorem, we get an 

x = c in (a, b) such that h'(c) = 0

Now,

h'(x) = f'(x) - [ f(b) - f(a) ] / (b - a)

h'(c) = f'(c) - [ f(b) - f(a) ] / (b - a) = 0

f'(c) = [ f(b) - f(a) ] / (b - a)

Thus the Mean Value Theorem is verified.

This result is true only for the function f(x) which is continuous and differentiable on the interval (a, b)

Graphical Representation of Mean Value Theorem

Mean Value Theorem can be easily interpreted using its graphical representation. Graphically mean value theorem state that for any function f(x) which is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) there exists a point 'c' in the closed [a, b] such that at [f(c), c] point on the curve, the tangent at that point is parallel to the secant drawn between the initial two points [a, f(a)] and [b, f(b)].

We know that,

Slope at any point c on the curve f(x) = f'(c)

Slope of the line passing through [a, f(a)] and [b, f(b)] = [ f(b) - f(a) ] / (b - a)

Now according to the mean value theorem graphical representation

Slope of the Tangent = Slope of the Secant

f'(c) = [f(b) - f(a)] / (b - a)

Difference Between Mean Value Theorem and Rolle's Theorem

Both the mean value theorem and Rolle's theorem define the function f(x) such that it is continuous across the interval [a, b], and it is differentiable across the interval (a, b). In the mean value theorem, the two referred points (a, f(a)), (b, f(b)) are distinct and f(a) ≠ f(b). In Rolle's theorem, the points are defined such that f(a) = f(b).

The value of c in the mean value theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the secant joining the two points. The value of c in Rolle's theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the x-axis. The slope in the mean value theorem is f'(c) = [ f(b) - f(a) ] / (b - a), and the slope in Rolle's theorem is equal to f'(c) = 0.

Learn more about, Rolle’s Theorem and Lagrange’s Mean Value Theorem

Cauchy Mean Value Theorem

Cauchy Mean Value Theorem is another important mean value theorem which is the extension of the mean value theorem.

Statement:

Let, f(x) and g(x) are the functions defined on [a, b] such that, f(x) and g(x) both are continuous on [a, b] and f(x) and g(x) both are differentiable on (a, b), then there exists at least one point c ∈ (a, b) such that

f'(c) / g'(c) = { f(b) - f(a)} / {g(b) - g(a)}

OR

f'(c) {g(b) - g(a)}= g'(c){ f(b) - f(a)}

This the Cauchy Mean Value Theorem.

If we take, g(x) = x for every x ∈ [a, b] in the above theorem, we get

f'(c) = { f(b) - f(a)} / {b-a}

This is the statement of Lagrange’s mean value theorem.

Generalized Mean Value Theorem

The generalized form of Mean value theorem is given as: For any three functions f, g, and h such that,

• f, g, and h are continuous in [a, b]
• f, g, and h are derivable in (a, b)

D(x) = \begin{vmatrix}f(x) &g(x)& h(x) \\ f(a) &g(a) &h(a) \\ f(b) &g(b) &h(b) \end{vmatrix}

\begin{vmatrix}f'(c) &g'(c)& h'(c) \\ f(a) &g(a) &h(a) \\ f(b) &g(b) &h(b) \end{vmatrix} = 0

Applications of Mean Value Theorem

Mean value theorem provides a relation between a function and its derivative and this theorem is widely used in mathematical concepts. The mean value is represented as,

Let f(x) be a continuous function in the closed interval [a, b] and f(x) is differentiable on the open interval (a, b), then

f'(c) = [f(b) - f(a)]/(b-a) c ∈ (a, b)

Also, if f(b) = f(a) then,

f(c) = 0, c ∈ (a, b)

Various applications of the mean value theorem are,

• For any two functions f(x) and g(x) if they are continuous in interval [a, b] and differentiable on the interval (a, b),

f'(c) = g'(c), c ∈ (a, b), then f(x) – g(x) is constant in [a, b]

• It is used in physical interpretation (like speed analysis).

• If the derivative of the function f(x) is greater than zero then f(x) is strictly an Increasing function.

• If the derivative of the function f(x) is less than zero then f(x) is strictly a decreasing function.

• Taylor Series and Number Theory use Mean Value Theorem

Increasing and Decreasing Function

With the help of the mean value theorem, we can find whether a function is increasing or decreasing.

Increasing Function

The function y = f(x) is said to be an increasing function of x in the interval (a, b). If the derivative of the f(x) is positive in the interval (a, b) i.e.

f'(x) > 0 for a x ∈ (a, b)

Decreasing Function

The function y = f(x) is said to be a decreasing function of x in the interval (a, b). If the derivative of the f(x) is negative in the interval (a, b) i.e.

f'(x) < 0 for a x ∈ (a, b)

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Mean Value Theorem Examples

Example 1: f(x) = (x - 1)(x - 2)(x - 3), x ∈ [0, 4]. Find c if the Mean value theorem is applied to f(x)

Solution:

f(x) = (x - 1)(x - 2)(x - 3),

x ∈ [0, 4]

f(x) = x³ - 6x² + 11x - 6

As f(x) is Polynomial in x

• f(x) is continuous on [0. 4]
• f(x) is differentiable on (0, 4)

Thus, all the conditions of Mean Value theorem are satisfied.

To verify theorem we have to find c ∈ (0, 4) such that

f’(c) = (f(4) - f(0))/(4 - 0)                                …(1)

Now, f(4) = (4 - 1)(4 - 2)(4 - 3) = 6

f(0) = (0 - 1)(0 - 2)(0 - 3) = - 6 and

f'(x) = 3x² - 12x + 11

f'(c) = 3c² - 12c + 11

(f(4) - f(0))/(4 - 0) = 3c² - 12c + 11         [From 1]

3 = 3c² - 12c + 11

3c² - 12c + 8 = 0

c = 2 ± 2/√3

Both values of c lie between 0 and 4.

Example 2: Verify LaGrange's mean value theorem for the function f(x) = log(x) on [1, e]

Solution: 

f(x) = log(x) 

f(x) is continuous on [1, e]

f(x) is differentiable on (1, e), Thus all the conditions of Mean Value theorem are satisfied.

We want to find c ∈ (1, e) such that 

f(e) - f(1) = f'(c) {e - 1}           … from (1) [By mean value theorem]

log e -log 1 = f'(c) {e - 1}   

1 - 0 = f'(c) {e - 1}   

f'(c) = 1/(e - 1)                      …[2]

Now,

f(x) = log(x)     

Hence, f'(x) = 1/x

f'(c) = 1/c           …[3]

From [2] and [3]      

1/(e - 1) = 1/c     

Hence,     

c = e - 1 which is lies in interval (1, e) 

Thus, Lagrange's mean value theorem verified.

Example 3: Test whether the following function is increasing or not. f(x) = x³ - 3x² + 3x - 100, x ∈ R

Solution:    

f(x) = x³ - 3x² + 3x - 100, x ∈ R

Differentiating with respect to x

f'(x) = 3x² - 6x + 3

       = 3(x - 1)²

Since (x - 1)² is always positive x ≠1

f'(x) >0, for all x ∈ R.

Hence, f(x) is an increasing function,  for all x ∈ R.

Example 4: Find the value of x for which the function f(x) = x³+12x²+36x+6,  x is increasing.   

Solution:

f(x) = x³ + 12x² + 36x + 6

f'(x) = 3x² + 24x + 36 = 3(x + 2)(x + 6)

Now, f'(x) > 0, as f(x) is increasing

3(x + 2)(x + 6) > 0

The above condition is true in two cases

Case 1: Both multiples are positive

x + 2 > 0 and x + 6 > 0

x > -2 and x > -6

x ∈ (-2, ∞)

Case 2: Both multiples are negative

x + 2 < 0 and x + 6 > 0

x < -2 and x < -6

x ∈ (-∞, -6) 

Thus, f(x) = x³ + 12x² + 36x + 6 is increasing in the interval, x∈ (-∞, -6) U (-2, ∞)

Example 5: For the function f(x) = 2x³ + 2, find all the values of c that satisfy the mean value theorem, in the interval [-2, 2].

Solution:

f(x) = 2x³ + 2 is a polynomial function and hence it is continuous and differentiable over the interval [-2, 2]

f'(x) = 6x

f(2) =2(2)³ + 2 = 18

f(-2) = 2(-2)³ + 2 = -14

f'(c) = [ f(2) - f(-2)] / (2 - (-2)) = [18 - (-14)] /4 =8

Let us find c in (-2, 2) such that f'(c) = 8

f'(c) = 6c

According to Mean Value Theorem

6c = 8

c = 8/6 = 1.33

Thus, c belongs to the interval [-2, 2].

Practice Questions - Mean Value Theorem

Q1. Find the value of c guaranteed by the Mean Value Theorem for the function f(x) = 2x² - 3x + 1 on the interval [1, 3].

Q2. Find the value of c guaranteed by the Mean Value Theorem for the function g(x) =x on the interval [1, 9].

Q3. Find the value of c guaranteed by the Mean Value Theorem for the function h(x) = ln⁡x on the interval [1, e].

Q4. Find the value of c guaranteed by the Mean Value Theorem for the function f(x) = x³- 3x on the interval [-1, 2].

Summary

The Mean Value Theorem (MVT) is a fundamental result in calculus that provides a link between the average rate of change of a function and its instantaneous rate of change.

This result is useful for understanding the behavior of functions and is often applied in various fields to find specific points where certain conditions hold, such as in optimization problems and analyzing function behavior. The MVT essentially guarantees that for smooth functions, the average slope between two points is achieved at some point within the interval.