Which Of The Following Functions Satisfy The Hypotheses Of The Mean Value Theorem On The Given Interval? Be Sure You Can Explain Why.A. $f(x)=\begin{cases} \frac{\sin (3 X)}{x} & 0\ \textless \ X \leq \pi \ 0 & X=0 \end{cases} \text{ On } [0,

Introduction

The Mean Value Theorem (MVT) is a fundamental concept in calculus that provides a powerful tool for analyzing the behavior of functions. It states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). In this article, we will explore which functions satisfy the hypotheses of the Mean Value Theorem on a given interval.

Hypotheses of the Mean Value Theorem

The Mean Value Theorem has two main hypotheses:

1. Continuity: The function f(x) must be continuous on the closed interval [a, b].
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b).

**Function A: $f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases} \text{ on } [0, \pi]$

To determine if Function A satisfies the hypotheses of the Mean Value Theorem, we need to analyze its continuity and differentiability on the given interval.

Continuity of Function A

Function A is defined as:

$$f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the function is defined as 0, which is a constant. Therefore, the function is continuous at x = 0.

For x > 0, the function is a quotient of two functions, sin(3x) and x. Since sin(3x) is continuous and x is continuous, the quotient is also continuous.

However, we need to check if the function is continuous at x = π. As x approaches π from the left, the function approaches 0. As x approaches π from the right, the function also approaches 0. Therefore, the function is continuous at x = π.

Differentiability of Function A

To determine if Function A is differentiable, we need to find its derivative.

Using the quotient rule, we get:

$$f'(x)=\begin{cases} \frac{3\cos(3x)}{x} - \frac{\sin(3x)}{x^2} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the derivative is 0, which is a constant. Therefore, the function is differentiable at x = 0.

For x > 0, the derivative is a quotient of two functions, 3cos(3x) and x, and sin(3x) and x^2. Since 3cos(3x) and x are continuous, and sin(3x) and x^2 are continuous, the quotient is also continuous.

However, we need to check if the derivative is continuous at x = π. As x approaches π from the left, the derivative approaches 0. As x approaches π from the right, the derivative also approaches 0. Therefore, the derivative is continuous at x = π.

Conclusion

Function A satisfies the hypotheses of the Mean Value Theorem on the interval [0, π]. The function is continuous and differentiable on the entire interval, and the derivative is continuous at the endpoints.

**Function B: $f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases} \text{ on } [0, 2\pi]$

To determine if Function B satisfies the hypotheses of the Mean Value Theorem, we need to analyze its continuity and differentiability on the given interval.

Continuity of Function B

Function B is defined as:

$$f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the function is defined as 0, which is a constant. Therefore, the function is continuous at x = 0.

For x > 0, the function is a quotient of two functions, sin(3x) and x. Since sin(3x) is continuous and x is continuous, the quotient is also continuous.

However, we need to check if the function is continuous at x = π. As x approaches π from the left, the function approaches 0. As x approaches π from the right, the function also approaches 0. Therefore, the function is continuous at x = π.

We also need to check if the function is continuous at x = 2π. As x approaches 2π from the left, the function approaches 0. As x approaches 2π from the right, the function also approaches 0. Therefore, the function is continuous at x = 2π.

Differentiability of Function B

To determine if Function B is differentiable, we need to find its derivative.

Using the quotient rule, we get:

$$f'(x)=\begin{cases} \frac{3\cos(3x)}{x} - \frac{\sin(3x)}{x^2} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the derivative is 0, which is a constant. Therefore, the function is differentiable at x = 0.

For x > 0, the derivative is a quotient of two functions, 3cos(3x) and x, and sin(3x) and x^2. Since 3cos(3x) and x are continuous, and sin(3x) and x^2 are continuous, the quotient is also continuous.

However, we need to check if the derivative is continuous at x = π. As x approaches π from the left, the derivative approaches 0. As x approaches π from the right, the derivative also approaches 0. Therefore, the derivative is continuous at x = π.

We also need to check if the derivative is continuous at x = 2π. As x approaches 2π from the left, the derivative approaches 0. As x approaches 2π from the right, the derivative also approaches 0. Therefore, the derivative is continuous at x = 2π.

Conclusion

Function B satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2π]. The function is continuous and differentiable on the entire interval, and the derivative is continuous at the endpoints.

**Function C: $f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases} \text{ on } [0, \pi/2]$

To determine if Function C satisfies the hypotheses of the Mean Value Theorem, we need to analyze its continuity and differentiability on the given interval.

Continuity of Function C

Function C is defined as:

$$f(x)=\begin{cases} \frac{\sin (3 x)}{x} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the function is defined as 0, which is a constant. Therefore, the function is continuous at x = 0.

For x > 0, the function is a quotient of two functions, sin(3x) and x. Since sin(3x) is continuous and x is continuous, the quotient is also continuous.

However, we need to check if the function is continuous at x = π/2. As x approaches π/2 from the left, the function approaches 0. As x approaches π/2 from the right, the function also approaches 0. Therefore, the function is continuous at x = π/2.

Differentiability of Function C

To determine if Function C is differentiable, we need to find its derivative.

Using the quotient rule, we get:

$$f'(x)=\begin{cases} \frac{3\cos(3x)}{x} - \frac{\sin(3x)}{x^2} & 0\ \textless \ x \leq \pi \\ 0 & x=0 \end{cases}$$

At x = 0, the derivative is 0, which is a constant. Therefore, the function is differentiable at x = 0.

For x > 0, the derivative is a quotient of two functions, 3cos(3x) and x, and sin(3x) and x^2. Since 3cos(3x) and x are continuous, and sin(3x) and x^2 are continuous, the quotient is also continuous.

However, we need to check if the derivative is continuous at x = π/2. As x approaches π/2 from the left, the derivative approaches 0. As x approaches π/2 from the right, the derivative also approaches 0. Therefore, the derivative is continuous at x = π/2.

Q&A: Mean Value Theorem

Q: What is the Mean Value Theorem?

A: The Mean Value Theorem is a fundamental concept in calculus that provides a powerful tool for analyzing the behavior of functions. It states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).

Q: What are the hypotheses of the Mean Value Theorem?

A: The Mean Value Theorem has two main hypotheses:

1. Continuity: The function f(x) must be continuous on the closed interval [a, b].
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b).

Q: What is the significance of the Mean Value Theorem?

A: The Mean Value Theorem has several important implications:

1. Existence of a maximum or minimum: If a function is continuous on a closed interval and differentiable on the open interval, then there exists a point where the function has a maximum or minimum value.
2. Rolle's Theorem: If a function is continuous on a closed interval and differentiable on the open interval, then there exists a point where the function has a derivative equal to zero.
3. L'Hopital's Rule: If a function is continuous on a closed interval and differentiable on the open interval, then the limit of the function as x approaches a point can be found using L'Hopital's Rule.

Q: How do I apply the Mean Value Theorem?

A: To apply the Mean Value Theorem, follow these steps:

1. Check the hypotheses: Verify that the function is continuous on the closed interval and differentiable on the open interval.
2. Find the derivative: Find the derivative of the function using the quotient rule or other differentiation rules.
3. Find the point c: Find the point c in the open interval where the derivative is equal to the average rate of change of the function.
4. Verify the result: Verify that the point c satisfies the equation f'(c) = (f(b) - f(a)) / (b - a).

Q: What are some common mistakes to avoid when applying the Mean Value Theorem?

A: Some common mistakes to avoid when applying the Mean Value Theorem include:

1. Not checking the hypotheses: Failing to verify that the function is continuous and differentiable on the given interval.
2. Not finding the derivative: Failing to find the derivative of the function using the correct differentiation rules.
3. Not finding the point c: Failing to find the point c in the open interval where the derivative is equal to the average rate of change of the function.
4. Not verifying the result: Failing to verify that the point c satisfies the equation f'(c) = (f(b) - f(a)) / (b - a).

Q: What are some real-world applications of the Mean Value Theorem?

A: The Mean Value Theorem has several real-world applications, including:

1. Optimization: The Mean Value Theorem can be used to find the maximum or minimum value of a function.
2. Physics: The Mean Value Theorem can be used to model the motion of an object under the influence of a force.
3. Economics: The Mean Value Theorem can be used to model the behavior of a company's revenue or cost.
4. Computer Science: The Mean Value Theorem can be used to analyze the behavior of algorithms and data structures.

Conclusion

The Mean Value Theorem is a powerful tool for analyzing the behavior of functions. By understanding the hypotheses and applying the theorem correctly, you can use it to find the maximum or minimum value of a function, model the motion of an object, or analyze the behavior of algorithms and data structures. Remember to check the hypotheses, find the derivative, find the point c, and verify the result to ensure that you are applying the Mean Value Theorem correctly.