Mean Value Theorem (MVT) Explained: Definition, Properties & Solved Examples

The Mean Value Theorem (MVT) states that for any continuous and differentiable function over a given interval, there exists at least one point where the instantaneous rate of change (the derivative) is equal to the average rate of change across the interval. This theorem provides valuable insights into the behavior of functions and serves as a critical tool in calculus for analyzing and predicting function dynamics.

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1. Introduction to the Mean Value Theorem:

The Mean Value Theorem (MVT) is one of the most fundamental theorems in calculus. It establishes a formal relationship between the derivative of a function and its average rate of change over an interval. This theorem bridges differential and integral calculus, helping to describe how a continuous function behaves across an interval. Whether you’re a math student or a professional using calculus in real-world scenarios, understanding the Mean Value Theorem is crucial for analyzing and predicting function behavior.

2. What is Mean Value Theorem:

The Mean Value Theorem (MVT) states that for a continuous function $f(x)$, which is differentiable on the open interval $(a, b)$ and continuous on the closed interval $[a, b]$, there exists at least one point $c$ in the interval $(a, b)$ such that the derivative of the function at $c$ is equal to the average rate of change of the function over that interval. Formally: $f'(c) = \dfrac{f(b) - f(a)}{b - a}$

This means that at some point in the interval, the instantaneous rate of change (derivative) matches the average rate of change over the entire interval.

3. Graphical Representation of Mean Value Theorem:

Graphically, the Mean Value Theorem can be visualized by drawing the secant line between points $(a, f(a))$ and $(b, f(b))$. The theorem guarantees at least one point $c$ in the interval $(a, b)$ where the slope of the tangent line to the curve $f(x)$ is parallel to the secant line. In other words, the slope of the tangent line at $c$ matches the average slope over the entire interval.

4. Difference Between Mean Value Theorem and Rolle's Theorem:

Although Mean Value Theorem (MVT) and Rolle's Theorem are closely related, they differ in their conditions and implications:

• Rolle's Theorem: States that if a function is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$. It deals specifically with functions that have equal values at the endpoints.

• Mean Value Theorem (MVT): Generalizes Rolle's Theorem by removing the condition $f(a) = f(b)$. Instead, it guarantees the existence of a point $c$ where the instantaneous rate of change equals the average rate of change.

5. Properties of Mean Value Theorem:

Some important properties of the Mean Value Theorem include:

• Existence of a Tangent: There is always at least one point in the interval where the tangent line's slope equals the average change rate.

• Function Continuity and Differentiability: The function must be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

• Average Rate of Change: The theorem directly connects the average rate of change of the function over an interval to the instantaneous rate of change at some point within that interval.

6. Mean Value Theorem Solved Examples:

Question: 1.

Applying the Mean Value Theorem to a Polynomial Function

Verify that the Mean Value Theorem applies to the function $f(x) = x^2 - 4x 3$ on the interval $[1, 5]$, and find the value of $c$ that satisfies the theorem.

Solution:

Step 1: Check if the function meets the conditions
$f(x) = x^2 - 4x 3$ is a polynomial, so it is continuous and differentiable on all real numbers, including the interval $[1, 5]$. Hence, the Mean Value Theorem can be applied.

Step 2: Find the average rate of change
The average rate of change of $f(x)$ on $[1, 5]$ is: $\dfrac{f(5) - f(1)}{5 - 1}$

First, calculate $f(5)$ and $f(1)$: $\begin{matrix} f(5) = 5^2 - 4(5) 3 = 25 - 20 3 = 8 \\ f(1) = 1^2 - 4(1) 3 = 1 - 4 3 = 0 \end{matrix}$

Now, calculate the average rate of change: $\dfrac{f(5) - f(1)}{5 - 1} = \dfrac{8 - 0}{4} = 2$

Step 3: Find the derivative of $f(x)$
The derivative of $f(x) = x^2 - 4x 3$ is: $f'(x) = 2x - 4$

Step 4: Set $f'(c) = 2$
According to the Mean Value Theorem, there exists a point $c \in (1, 5)$ such that: $f'(c) = 2$

Set $2c - 4 = 2$: $\space 2c - 4 = 2 \rArr 2c = 6 \rArr c = 3$

Answer:
The value of $c$ that satisfies the Mean Value Theorem is $c = 3$.

Question: 2.

Applying the Mean Value Theorem to a Trigonometric Function

Verify that the Mean Value Theorem applies to the function $f(x) = \sin(x)$ on the interval $[0, \pi]$, and find the value of $c$ that satisfies the theorem.

Solution:

Step 1: Check if the function meets the conditions

• The function $f(x) = \sin(x)$ is continuous and differentiable on the interval $[0, \pi]$, so the Mean Value Theorem can be applied.

Step 2: Find the average rate of change
The average rate of change of $f(x) = \sin(x)$ on $[0, \pi]$ is: $\dfrac{f(\pi) - f(0)}{\pi - 0}$

Calculate $f(\pi)$ and $f(0)$: $\begin{matrix} f(\pi) = \sin(\pi) = 0 \\ f(0) = \sin(0) = 0 \end{matrix}$

Now, calculate the average rate of change: $\dfrac{f(\pi) - f(0)}{\pi - 0} = \dfrac{0 - 0}{\pi} = 0$

Step 3: Find the derivative of $f(x)$
The derivative of $f(x) = \sin(x)$ is: $f'(x) = \cos(x)$

Step 4: Set $f'(c) = 0$
According to the Mean Value Theorem, there exists a point $c \in (0, \pi)$ such that: $f'(c) = 0$

Since $f'(x) = \cos(x)$, set $\cos(c) = 0$. This occurs when: $c = \dfrac{\pi}{2}$

Answer:
The value of $c$ that satisfies the Mean Value Theorem is $c = \dfrac{\pi}{2}$.

Question: 3.

Applying the Mean Value Theorem to a Rational Function

Verify that the Mean Value Theorem applies to the function $f(x) = \dfrac{1}{x}$ on the interval $[1, 4]$, and find the value of $c$ that satisfies the theorem.

Solution:

Step 1: Check if the function meets the conditions
The function $f(x) = \dfrac{1}{x}$ is continuous and differentiable on the interval $(1, 4)$ because it is defined and differentiable for all $x \neq 0$. Therefore, it satisfies the conditions of the Mean Value Theorem on $[1, 4]$.

Step 2: Find the average rate of change
The average rate of change of $f(x)$ on the interval $[1, 4]$ is: $\dfrac{f(4) - f(1)}{4 - 1}$

Calculate $f(4)$ and $f(1)$: $\space f(4) = \dfrac{1}{4}, \ f(1) = \dfrac{1}{1} = 1$

Now, calculate the average rate of change: $\space \dfrac{\dfrac{1}{4} - 1}{4 - 1} = \dfrac{-\dfrac{3}{4}}{3} = -\dfrac{1}{4}$

Step 3: Find the derivative of $f(x)$
The derivative of $f(x) = \dfrac{1}{x}$ is: $f'(x) = -\dfrac{1}{x^2}$

Step 4: Set $f'(c) = -\dfrac{1}{4}$
According to the Mean Value Theorem, there exists a point $c \in (1, 4)$ such that: $f'(c) = -\dfrac{1}{4}$

Set $-\dfrac{1}{c^2} = -\dfrac{1}{4}$: $\dfrac{1}{c^2} = \dfrac{1}{4}$

Solving for $c$: $c^2 = 4 \space \rArr \space c = 2$

Answer:
The value of $c$ that satisfies the Mean Value Theorem for $f(x) = \dfrac{1}{x}$ on $[1, 4]$ is $c = 2$.

Question: 4.

Applying the Mean Value Theorem to a Cubic Function

Verify that the Mean Value Theorem applies to the function $f(x) = x^3 - 3x^2 2x$ on the interval $[0, 3]$, and find the value of $c$ that satisfies the theorem.

Solution:

Step 1: Check if the function meets the conditions
The function $f(x) = x^3 - 3x^2 2x$ is a polynomial, which means it is continuous and differentiable on all real numbers, including the interval $[0, 3]$. Thus, the Mean Value Theorem can be applied.

Step 2: Find the average rate of change
The average rate of change of $f(x)$ on $[0, 3]$ is: $\dfrac{f(3) - f(0)}{3 - 0}$

Calculate $f(3)$ and $f(0)$: $\begin{matrix} f(3) = 3^3 - 3(3^2) 2(3) = 27 - 27 6 = 6 \\ f(0) = 0^3 - 3(0^2) 2(0) = 0 \end{matrix}$

Now, calculate the average rate of change: $\dfrac{f(3) - f(0)}{3 - 0} = \dfrac{6 - 0}{3} = 2$

Step 3: Find the derivative of $f(x)$
The derivative of $f(x) = x^3 - 3x^2 2x$ is: $f'(x) = 3x^2 - 6x 2$

Step 4: Set $f'(c) = 2$
According to the Mean Value Theorem, there exists a point $c \in (0, 3)$ such that: $f'(c) = 2$

Set $3c^2 - 6c 2 = 2$: $\space 3c^2 - 6c 2 = 2$

Simplify the equation: $3c^2 - 6c = 0$

Factor the equation: $3c(c - 2) = 0$

This gives two solutions: $c = 0$ or $c = 2$.

5. Step 5: Choose the value of $c$ in the open interval
   Since $c = 0$ is the endpoint of the interval, it is excluded by the Mean Value Theorem, which only applies to the open interval $(0, 3)$. Therefore, the only valid value is $c = 2$.

Answer:
The value of $c$ that satisfies the Mean Value Theorem for $f(x) = x^3 - 3x^2 2x$ on $[0, 3]$ is $c = 2$.

7. Practice Questions on Mean Value Theorem:

Q:1. Find the value of $c$ that satisfies the Mean Value Theorem for $f(x) = x^3$ on the interval $[1, 3]$.

Q:2. Apply the Mean Value Theorem to $f(x) = e^x$ on the interval $[0, 2]$.

Q:3. Verify the Mean Value Theorem for $f(x) = \ln(x)$ on $[1, e]$.

8. FAQs on Mean Value Theorem:

What is the Mean Value Theorem?

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there is at least one point $c \in (a, b)$ where the derivative of the function equals the average rate of change over the interval.

What is the significance of the Mean Value Theorem?

The MVT provides a formal relationship between the instantaneous rate of change and the average rate of change of a function, ensuring that there is at least one point where the two are equal.

What are the conditions for applying the Mean Value Theorem?

The function must be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

How does the Mean Value Theorem relate to Rolle’s Theorem?

Rolle’s Theorem is a special case of the MVT where $f(a) = f(b)$, meaning that the average rate of change is zero, and there exists at least one point where the derivative is also zero.

Can the Mean Value Theorem be applied to non-polynomial functions?

Yes, the MVT can be applied to any function that meets the conditions of continuity and differentiability, regardless of whether it’s a polynomial or not.

What is the graphical interpretation of the Mean Value Theorem?

Graphically, the MVT guarantees that there is at least one point on the curve where the tangent line is parallel to the secant line connecting $(a, f(a))$ and $(b, f(b))$.

Why is differentiability required for the Mean Value Theorem?

Differentiability ensures that the function has a well-defined slope (tangent line) at every point in the interval, which is crucial for the theorem’s application.

What happens if a function does not meet the conditions of the Mean Value Theorem?

If the function is not continuous or not differentiable, the MVT cannot be applied, and there is no guarantee of a point where the instantaneous rate of change equals the average rate of change.

9. Real-life Application of Mean Value Theorem:

The Mean Value Theorem has practical applications in physics, particularly in motion analysis. For instance, if an object moves from one point to another, the Mean Value Theorem guarantees at least one point during its journey where its instantaneous velocity matches the average velocity over the entire trip. It is also applied in economics to estimate average rates of change in data trends and in engineering to solve optimization problems.

10. Conclusion:

The Mean Value Theorem is a cornerstone concept in calculus that connects a function's average rate of change to its instantaneous rate of change. The theorem provides valuable insights into the behavior of continuous and differentiable functions by guaranteeing that at least one point in the interval matches the average rate. Whether you’re solving problems in physics, economics, or engineering, understanding and applying the Mean Value Theorem is essential.

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