The Goal Of This Problem Is To Show That The Function F ( X ) = 1 − 6 X 2 F(x) = 1 - 6x^2 F ( X ) = 1 − 6 X 2 Satisfies Both The Conditions (the Hypotheses) And The Conclusion Of The Mean Value Theorem For X X X In The Interval − 3 , 5 {-3, 5} − 3 , 5 .Verification Of

by ADMIN 269 views

Introduction

The Mean Value Theorem (MVT) is a fundamental concept in calculus that provides a powerful tool for analyzing functions and their behavior. 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 the application of the MVT to the function f(x) = 1 - 6x^2 and verify that it satisfies both the conditions and the conclusion of the MVT for x in the interval [-3, 5].

The Function and Its Properties

The given function is f(x) = 1 - 6x^2. To apply the MVT, we need to check if the function satisfies the conditions of continuity and differentiability on the given interval [-3, 5].

Continuity of the Function

A function f(x) is said to be continuous at a point x = a if the following conditions are satisfied:

  1. The function is defined at x = a.
  2. The limit of the function as x approaches a exists.
  3. The limit of the function as x approaches a is equal to the value of the function at x = a.

In this case, the function f(x) = 1 - 6x^2 is a polynomial function, which is continuous everywhere. Therefore, the function is continuous on the interval [-3, 5].

Differentiability of the Function

A function f(x) is said to be differentiable at a point x = a if the following conditions are satisfied:

  1. The function is continuous at x = a.
  2. The limit of the difference quotient (f(x) - f(a)) / (x - a) exists as x approaches a.

In this case, the function f(x) = 1 - 6x^2 is a polynomial function, which is differentiable everywhere. Therefore, the function is differentiable on the interval [-3, 5].

Verification of the Conditions

Now that we have established the continuity and differentiability of the function f(x) = 1 - 6x^2 on the interval [-3, 5], we can proceed to verify that it satisfies the conditions of the MVT.

Condition 1: Continuity on the Closed Interval

As we have already established, the function f(x) = 1 - 6x^2 is continuous on the interval [-3, 5].

Condition 2: Differentiability on the Open Interval

As we have already established, the function f(x) = 1 - 6x^2 is differentiable on the interval [-3, 5].

Condition 3: Existence of a Point c in the Open Interval

The MVT states that there exists a point c in the open interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). In this case, we need to find a point c in the interval (-3, 5) such that f'(c) = (f(5) - f(-3)) / (5 - (-3)).

Calculation of f'(c)

To find the derivative of the function f(x) = 1 - 6x^2, we can use the power rule of differentiation, which states that if f(x) = x^n, then f'(x) = nx^(n-1).

In this case, we have f(x) = 1 - 6x^2, so f'(x) = -12x.

Calculation of f(5) and f(-3)

To find the values of f(5) and f(-3), we can substitute x = 5 and x = -3 into the function f(x) = 1 - 6x^2.

f(5) = 1 - 6(5)^2 = 1 - 150 = -149

f(-3) = 1 - 6(-3)^2 = 1 - 54 = -53

Calculation of f'(c)

Now that we have found the values of f(5) and f(-3), we can substitute them into the equation f'(c) = (f(5) - f(-3)) / (5 - (-3)).

f'(c) = (-149 - (-53)) / (5 - (-3)) = (-149 + 53) / (5 + 3) = -96 / 8 = -12

Verification of the Conclusion

The MVT states that f'(c) = (f(b) - f(a)) / (b - a). In this case, we have found that f'(c) = -12, which is equal to (f(5) - f(-3)) / (5 - (-3)).

Therefore, we have verified that the function f(x) = 1 - 6x^2 satisfies the conditions and the conclusion of the MVT for x in the interval [-3, 5].

Conclusion

In this article, we have applied the Mean Value Theorem to the function f(x) = 1 - 6x^2 and verified that it satisfies both the conditions and the conclusion of the MVT for x in the interval [-3, 5]. We have established the continuity and differentiability of the function on the given interval and found a point c in the open interval such that f'(c) = (f(b) - f(a)) / (b - a). The results of this analysis demonstrate the power and versatility of the MVT in analyzing functions and their behavior.

References

  • [1] Thomas, G. B. (2010). Calculus and Analytic Geometry. Addison-Wesley.
  • [2] Stewart, J. (2012). Calculus: Early Transcendentals. Cengage Learning.
  • [3] Anton, H. (2012). Calculus: A New Horizon. John Wiley & Sons.

Glossary

  • Mean Value Theorem (MVT): A fundamental concept in calculus that provides a powerful tool for analyzing functions and their behavior.
  • Continuity: A function f(x) is said to be continuous at a point x = a if the following conditions are satisfied: the function is defined at x = a, the limit of the function as x approaches a exists, and the limit of the function as x approaches a is equal to the value of the function at x = a.
  • Differentiability: A function f(x) is said to be differentiable at a point x = a if the following conditions are satisfied: the function is continuous at x = a, and the limit of the difference quotient (f(x) - f(a)) / (x - a) exists as x approaches a.
  • Derivative: The derivative of a function f(x) is denoted by f'(x) and represents the rate of change of the function with respect to x.
    The Mean Value Theorem: A Comprehensive Q&A Guide =====================================================

Introduction

The Mean Value Theorem (MVT) is a fundamental concept in calculus that provides a powerful tool for analyzing functions and their behavior. In our previous article, we explored the application of the MVT to the function f(x) = 1 - 6x^2 and verified that it satisfies both the conditions and the conclusion of the MVT for x in the interval [-3, 5]. In this article, we will provide a comprehensive Q&A guide to help you better understand the MVT and its applications.

Q: What is the Mean Value Theorem?

A: The Mean Value Theorem (MVT) is a fundamental concept in calculus that 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 conditions for the Mean Value Theorem?

A: The conditions for the Mean Value Theorem are:

  1. The function f(x) must be continuous on the closed interval [a, b].
  2. The function f(x) must be differentiable on the open interval (a, b).
  3. The point c must be in the open interval (a, b).

Q: How do I apply the Mean Value Theorem?

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

  1. Check if the function f(x) is continuous on the closed interval [a, b].
  2. Check if the function f(x) is differentiable on the open interval (a, b).
  3. Find the derivative of the function f(x).
  4. Find the values of f(a) and f(b).
  5. Substitute the values of f(a), f(b), and f'(c) into the equation f'(c) = (f(b) - f(a)) / (b - a) to find the value of c.

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

A: The Mean Value Theorem is significant because it provides a powerful tool for analyzing functions and their behavior. It can be used to:

  1. Find the maximum and minimum values of a function.
  2. Determine the intervals where a function is increasing or decreasing.
  3. Find the points of inflection of a function.
  4. Analyze the behavior of a function over a given interval.

Q: What are some common applications of the Mean Value Theorem?

A: Some common applications of the Mean Value Theorem include:

  1. Finding the maximum and minimum values of a function.
  2. Determining the intervals where a function is increasing or decreasing.
  3. Finding the points of inflection of a function.
  4. Analyzing the behavior of a function over a given interval.
  5. Solving optimization problems.

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. Failing to check if the function is continuous on the closed interval [a, b].
  2. Failing to check if the function is differentiable on the open interval (a, b).
  3. Failing to find the derivative of the function.
  4. Failing to find the values of f(a) and f(b).
  5. Failing to substitute the values of f(a), f(b), and f'(c) into the equation f'(c) = (f(b) - f(a)) / (b - a).

Conclusion

In this article, we have provided a comprehensive Q&A guide to help you better understand the Mean Value Theorem and its applications. We have covered the conditions for the Mean Value Theorem, how to apply it, and some common applications and mistakes to avoid. By following these guidelines, you will be able to apply the Mean Value Theorem with confidence and solve a wide range of problems in calculus.

References

  • [1] Thomas, G. B. (2010). Calculus and Analytic Geometry. Addison-Wesley.
  • [2] Stewart, J. (2012). Calculus: Early Transcendentals. Cengage Learning.
  • [3] Anton, H. (2012). Calculus: A New Horizon. John Wiley & Sons.

Glossary

  • Mean Value Theorem (MVT): A fundamental concept in calculus that provides a powerful tool for analyzing functions and their behavior.
  • Continuity: A function f(x) is said to be continuous at a point x = a if the following conditions are satisfied: the function is defined at x = a, the limit of the function as x approaches a exists, and the limit of the function as x approaches a is equal to the value of the function at x = a.
  • Differentiability: A function f(x) is said to be differentiable at a point x = a if the following conditions are satisfied: the function is continuous at x = a, and the limit of the difference quotient (f(x) - f(a)) / (x - a) exists as x approaches a.
  • Derivative: The derivative of a function f(x) is denoted by f'(x) and represents the rate of change of the function with respect to x.