Mean Value Theorem

● Mean Value Theorem

● What is MVT Theorem?
● How to Represent Mean Value Theorem Graphically
● Prooving Mean Value Theorem
● Mean Value Theorem Examples:
● General Use of Mean Value Theorem
● Practice problems
● Frequently Asked Question About Mean Value Theurem

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental theorem in calculus that establishes a relationship between
the average rate of change and the instantaneous rate of change of a function over a closed interval. The
theorem is often attributed to the French mathematician Augustin-Louis Cauchy and is a key result in the study
of calculus.

This theorem essentially guarantees the existence of a point where the instantaneous rate of change (slope of
the tangent line) is equal to the average rate of change (slope of the secant line) over the given interval. The
MVT Theorem is a crucial tool in calculus and is used to prove other theorems and results.

In mathematical terms, if \( f(x) \) satisfies the conditions mentioned above, then there exists a \( c \) in \((a, b)\)
such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

where \( f'(c) \) represents the derivative of \( f(x) \) evaluated at \( c \).

This theorem essentially guarantees the existence of a point where the instantaneous rate of change (slope of
the tangent line) is equal to the average rate of change (slope of the secant line) over the given interval. The
MVT Theorem is a crucial tool in calculus and is used to prove other theorems and results.

What is MVT Theorem?

The Mean Value Theorem is a fundamental thoerem in calculus that states the following:

If a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a,
b)\), then there exists at least one number \( c \) in the open interval \((a, b)\) such that the instantaneous rate
of change (derivative) of the function at \( c \) is equal to the average rate of change of the function over the
interval \([a, b]\).

In mathematical terms, if \( f(x) \) satisfies the conditions mentioned above, then there exists a \( c \) in \((a, b)\)
such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

where,
\(f'(c)\) represents the derivative of the function \(f(x)\) evaluated at the point \(c\).
\(f(b)\) and \(f(a)\) are the values of the function at the endpoints of the interval.
\(b\) and \(a\) are the endpoints of the interval.
The Mean Value Theorem is a powerful tool in calculus, and it has important consequences and applications in
the study of functions and their behavior.

How to Represent Mean Value Theorem Graphically

The Mean Value Theorem (MVT) can be visually understood through a graphical representation. Here's a
description of how the theorem works graphically:

Scenario:
Consider a function \(f(x)\) that is continuous on the closed interval \([a, b]\) and differentiable on the open
interval \((a, b)\). The Mean Value Theorem states that there exists at least one point \(c\) in \((a, b)\) where the
instantaneous rate of change (slope of the tangent line) is equal to the average rate of change (slope of the
secant line) over the interval \([a, b]\).

Graphical Interpretation:

1. Continuous and Differentiable Function:

The function \(f(x)\) is continuous, meaning there are no breaks or holes in its graph, and differentiable,
meaning it has a well-defined derivative for every point in the interval \((a, b)\).

2. Secant Line:
Draw the secant line that connects the points \(A (a, f(a))\) and \(B (b, f(b))\). This line represents the average
rate of change of \(f\) over the interval \([a, b]\).

3. Tangent Line:
The geometric interpretation of the Mean Value Theorem is that there exists at least one point \(c\) in the
interval \((a, b)\) where the tangent line to the graph of \(f\) at \(c\) is parallel to the secant line passing through
the points \(A (a, f(a))\) and \(B (b, f(b))\).In other words, there's a point where the instantaneous rate of
change equals the average rate of change.

4. Visual Representation:
On the graph, visualize the tangent line touching the curve at a specific point \(c\) such that the slope of the
tangent line equals the slope of the secant line.
The tangent line to f(x) at x = c is \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
The slope of the secant line connecting \(A (a, f(a))\) and \(B (b, f(b))\), using the slope formula is \( \frac{f(b) -
f(a)}{b - a} \).
As the slope values are the same, the tangent line is parallel to the secant line.

This graphical representation helps to understand the geometric interpretation of the Mean Value Theorem and
how it guarantees the existence of a specific point \(c\) with certain properties. The slopes of the tangent and
secant lines play a key role in the theorem's visual explanation.

Proving Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental theorem in calculus that describes the behavior of
derivatives within a given interval of a continuous function. Here's how the theorem is stated and a proof for it:

Statement of the Mean Value Theorem

Suppose \( f \) is a function that meets the following two criteria:
1. \( f \) is continuous on the closed interval \([a, b]\).
2. \( f \) is differentiable on the open interval \((a, b)\).

Then there exists at least one point \( c \) in the open interval \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

This essentially states that there is at least one point at which the derivative of the function \( f \) is equal to the
average rate of change (or slope) of the function over the interval \([a, b]\).

Proof of the Mean Value Theorem

1. Define an Auxiliary Function:
Let \(g(x) \) represent vertical distance between the point \( (x,f(x)) \) and the point \( (x,y)\) on that line.
Therefore,
\[ g(x) = f(x) - \left[ f(a) + \frac{f(b) - f(a)}{b - a} (x - a) \right] \]
This function \( g(x) \) subtracts the equation of the straight line connecting the points \((a, f(a))\) and \((b,
f(b))\) from \( f(x) \). Notice that:
\[ g(a) = f(a) - f(a) = 0 \]
\[ g(b) = f(b) - \left[ f(a) + \frac{f(b) - f(a)}{b - a} (b - a) \right] = f(b) - f(b) = 0 \]
So, \( g(a) = g(b) = 0 \).

2. Apply Rolle's Theorem:

Since \( g \) is defined in terms of \( f \) and the linear function, and \( f \) is continuous on \([a, b]\) and
differentiable on \((a, b)\), \( g \) also is continuous on \([a, b]\) and differentiable on \((a, b)\). By Rolle's
theorem, there exists some \( c \) in \((a, b)\) such that
\[ g'(c) = 0 \]

3. Differentiate \( g(x) \):

Differentiating \( g(x) \) gives:
\[ g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} \]
Setting \( g'(c) = 0 \) yields:
\[ f'(c) - \frac{f(b) - f(a)}{b - a} = 0 \]
Therefore:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

And this completes the proof of the Mean Value Theorem. The theorem confirms that at some point \( c \)
within the interval \((a, b)\), the instantaneous rate of change (derivative) of the function \( f \) matches the
average rate of change over the entire interval.
General Use of Mean Value Theorem
The Mean Value Theorem (MVT) has various applications across different areas of mathematics and science.
Here are some general uses of the Mean Value Theorem:

1. Physics: Motion and Velocity: In physics, the MVT is applied to describe motion and velocity. It ensures that
at some point in time, an object's instantaneous velocity is equal to its average velocity over an interval.

2. Economics: Average and Marginal Concepts: In economics, the MVT is used to connect average and
marginal concepts, such as average cost and marginal cost.

3. Engineering: Design and Optimization: Engineers use the MVT in designing systems and structures,
ensuring that certain conditions are met at specific points.

4. Computer Science: Algorithm Analysis: In computer science, the MVT may be used in algorithm analysis
to understand the rate of change of certain functions.

5. Statistics: Regression Analysis: The MVT has applications in statistics, especially in regression analysis,
where it helps interpret the relationship between variables.

6. Environmental Science: Modeling Environmental Processes: The MVT is employed in environmental

science to model various processes, such as the rate of change of pollutant concentrations.

7. Medicine: Drug Dosage Modeling: In medicine, the MVT can be used to model drug dosage, ensuring that
the concentration of a drug reaches a desired level.

8. Finance: Rate of Return: The MVT can be applied in finance to analyze the rate of return on investments
over specific time intervals.

These are just a few examples, and the Mean Value Theorem finds application wherever the rate of change of
a quantity is involved, making it a versatile tool in various scientific and mathematical disciplines.

Solved Examples

Example 1: Consider the function \(f(x) = x^2\) on the interval \([1, 3]\). We want to find a point \(c\) in
\((1, 3)\) such that \(f'(c)\) equals the average rate of change of \(f\) over \([1, 3]\).

Solution:

1. Average Rate of Change:

\[ \text{Average Rate of Change} = \frac{f(3) - f(1)}{3 - 1} \]

2. Derivative:
\[ f'(x) = 2x \]

3. Find \(c\):
Solve \(f'(c) = \frac{f(3) - f(1)}{3 - 1}\) for \(c\).

\[ 2c = \frac{9 - 1}{2} \]
 \[ 2c = 4 \]
\[ c = 2 \]

So, by the Mean Value Theorem, there exists at least one \(c\) in \((1, 3)\) such that \(f'(c)\) equals the average
rate of change.

Example 2: Consider the function \(g(x) = \sqrt{x}\) on the interval \([1, 4]\). We want to find a point \(c\)
in \((1, 4)\) such that \(g'(c)\) equals the average rate of change of \(g\) over \([1, 4]\).

Solution:
1. Average Rate of Change:
\[ \text{Average Rate of Change} = \frac{g(4) - g(1)}{4 - 1} \]

2. Derivative:
\[ g'(x) = \frac{1}{2\sqrt{x}} \]

3. Find \(c\):
Solve \(g'(c) = \frac{g(4) - g(1)}{4 - 1}\) for \(c\).

\[ \frac{1}{2\sqrt{c}} = \frac{2 - 1}{3} \]

\[ \frac{1}{2\sqrt{c}} = \frac{1}{3} \]
\[ 2\sqrt{c} = 3 \]
\[ \sqrt{c} = \frac{3}{2} \]
\[ c = \frac{9}{4} \]

So, by the Mean Value Theorem, there exists at least one \(c\) in \((1, 4)\) such that \(g'(c)\) equals the average
rate of change.

Example 3: Consider the function \(h(x) = e^x\) on the interval \([0, 2]\). Find a point \(c\) in \((0, 2)\)
where \(h'(c)\) equals the average rate of change of \(h\) over \([0, 2]\).

Solution:
Let's find a point \(c\) in \((0, 2)\) where \(h'(c)\) equals the average rate of change of \(h\) over \([0, 2]\).

The function \(h(x) = e^x\) has the derivative \(h'(x) = e^x\).

Average Rate of Change:

\[ \text{Average Rate of Change} = \frac{h(2) - h(0)}{2 - 0} = \frac{e^2 - 1}{2} \]

Derivative:
\[ h'(x) = e^x \]

Find \(c\):
Solve \(h'(c) = \frac{e^2 - 1}{2}\) for \(c\):
\[ e^c = \frac{e^2 - 1}{2} \]

Taking the natural logarithm of both sides:

\[ c = \ln\left(\frac{e^2 - 1}{2}\right) \]

So, \(c = \ln\left(\frac{e^2 - 1}{2}\right)\) is the point in \((0, 2)\) where \(h'(c)\) equals the average rate of change
of \(h\) over \([0, 2]\).

Example 4: Let \(p(x) = \ln(x)\) on the interval \([1, 5]\). Find a point \(c\) in \((1, 5)\) such that \(p'(c)\)
equals the average rate of change of \(p\) over \([1, 5]\).
Solution:
Let's find a point \(c\) in \((1, 5)\) where \(p'(c)\) equals the average rate of change of \(p\) over \([1, 5]\).

The function \(p(x) = \ln(x)\) has the derivative \(p'(x) = \frac{1}{x}\).

Average Rate of Change:

\[ \text{Average Rate of Change} = \frac{p(5) - p(1)}{5 - 1} = \frac{\ln(5) - \ln(1)}{4} = \frac{\ln(5)}{4} \]

Derivative:
\[ p'(x) = \frac{1}{x} \]

Find \(c\):
Solve \(p'(c) = \frac{\ln(5)}{4}\) for \(c\):
\[ \frac{1}{c} = \frac{\ln(5)}{4} \]

Taking the reciprocal of both sides:

\[ c = \frac{4}{\ln(5)} \]

So, \(c = \frac{4}{\ln(5)}\) is the point in \((1, 5)\) where \(p'(c)\) equals the average rate of change of \(p\) over
\([1, 5]\).

Example 5: Consider the function \(q(x) = \sin(x)\) on the interval \([0, \pi]\). Find a point \(c\) in \((0, \pi)\)
where \(q'(c)\) equals the average rate of change of \(q\) over \([0, \pi]\).

Solution:
Let's find a point \(c\) in \((0, \pi)\) where \(q'(c)\) equals the average rate of change of \(q\) over \([0, \pi]\).

The function \(q(x) = \sin(x)\) has the derivative \(q'(x) = \cos(x)\).

Average Rate of Change:

\[ \text{Average Rate of Change} = \frac{q(\pi) - q(0)}{\pi - 0} = \frac{\sin(\pi) - \sin(0)}{\pi} = 0 \]

Derivative:
\[ q'(x) = \cos(x) \]

Find \(c\):
Solve \(q'(c) = 0\) for \(c\):
\[ \cos(c) = 0 \]

The solutions for \(\cos(c) = 0\) occur when \(c = \frac{\pi}{2} + k\pi\), where \(k\) is an integer. Since \(c\) needs
to be in \((0, \pi)\), the solution is \(c = \frac{\pi}{2}\).

So, \(c = \frac{\pi}{2}\) is the point in \((0, \pi)\) where \(q'(c)\) equals the average rate of change of \(q\) over
\([0, \pi]\).

Practice Problems

Q1. Determine the value of \(c\) that satisfies the Mean Value Theorem for the function \(f(z) = z^2 - 4z +
3\) on the interval \([1, 4]\).

a. c=2.8
 b. c=3
c. c=7
d. c=2.5
Answer: d

Q2. Find the values of \(c\) that satisfy the Mean Value Theorem for the function \(f(y) = y^3 + 6\) on the
interval \([1, 3]\).

a. c= 4.3
b. c= 2.08
c. c=2.5
d. d=1.08
Answer: b

Q4. Find the values of \(c\) that satisfy the Mean Value Theorem for the function \(f(x) = z^2 - 5z + 7\) on
the interval \([-1, 3]\).

a. c=-1
b. c=1
c. c=0.5
d. c=-0.5
Answer: b

Q5. Determine the values of \(c\) that satisfy the Mean Value Theorem for the function \(f(x) = 4x^2\) on
the interval \([2, 5]\).

a. c=0.5
b. c=1.5
c. c=2.5
d. c=3.5
Answer: d

Frequently Asked Questions

Q1. What is the Mean Value Theorem (MVT)?

Answer: The Mean Value Theorem is a fundamental theorem in calculus that guarantees the existence of at
least one point in an interval where the instantaneous rate of change (derivative) of a function is equal to the
average rate of change over the entire interval.

Q2. What are the conditions for the Mean Value Theorem to apply?
Answer: The function must be continuous on the closed interval and differentiable on the open interval.

Q3. How is the Mean Value Theorem used in calculus?

Answer: The MVT is often used to prove other theorems and results in calculus. It is also employed to
understand and analyze the behavior of functions, especially when dealing with rates of change.

Q4. Can the Mean Value Theorem be applied to non-differentiable functions?

Answer: No, the Mean Value Theorem requires the function to be differentiable on the open interval. If the
function is not differentiable, the theorem does not apply.

Q5. What is the geometric interpretation of the Mean Value Theorem?

Answer: Geometrically, the MVT implies that there exists at least one point in the interval where the slope of
the tangent line (instantaneous rate of change) is equal to the slope of the secant line (average rate of
change).

Q6. Can the Mean Value Theorem be applied to a constant function?

Answer: Yes, the MVT can be applied to a constant function. In this case, the derivative is zero, and any point
in the interval satisfies the condition.

Q7. How is the Mean Value Theorem related to Rolle's Theorem?

Answer: Rolle's Theorem is a special case of the Mean Value Theorem. If a function is continuous on a closed
interval, differentiable on the open interval, and has equal function values at the endpoints, then there exists at
least one point in the interval where the derivative is zero.

Q8. What are some practical applications of the Mean Value Theorem?
Answer: The MVT is applied in physics, economics, engineering, and other fields to analyze rates of change,
motion, optimization, and related concepts.

Q9. Are there variations of the Mean Value Theorem?

Answer: Yes, there are variations such as the Generalized Mean Value Theorem and Cauchy's Mean Value
Theorem, each with its own conditions and statements.

Q10. Can the Mean Value Theorem be used to find the maximum or minimum values of a function?
Answer: No, the MVT does not directly provide information about maximum or minimum values. It provides
information about rates of change. The Extreme Value Theorem is more relevant for finding maximum and
minimum values.