1.16 Working with the Intermediate Value Theorem (IVT Calc)

Intermediate Value Theorem (IVT)

Objectives:

• Understand the concept of the Intermediate Value Theorem.
• Apply the Intermediate Value Theorem to find and prove the existence of roots for a function.
• Use the Intermediate Value Theorem to prove the existence of a solution to a problem.

Essential Knowledge:

• The Intermediate Value Theorem states that for any value c between the minimum and maximum values of a continuous function, there exists a point at which the function takes on the value c.
• The Intermediate Value Theorem can be used to find and prove the existence of roots for a function.
• The Intermediate Value Theorem can be used to prove the existence of a solution to a problem.

Overview:

Examples:

1. Given the function f(x) = x^2 - 2. We know that f(1) = -1 and f(2) = 2. Using the IVT, we can prove that there exists at least one root of the function between x = 1 and x = 2.
2. Given the function g(x) = x^3 - 6x^2 + 11x - 6. We know that g(-1) = -4 and g(1) = 4. Using the IVT, we can prove that there exists at least one root of the function between x = -1 and x = 1. By using the techniques of factoring, we can find that the roots are x=-2, x=-1 and x=1
3. Given the function h(x) = x^2+3x+2. We know that h(0) = 2 and h(1) = 4. Using the IVT, we can prove that there are no roots of the function between x = 0 and x = 1. By using the techniques of factoring, we can find that the roots are x=-1 and x=-2, which are not between 0 and 1.
4. Given the function j(x) = x^3-9x+3. We know that j(-1) = -7 and j(1) = -5. Using the IVT, we can prove that there exists at least one root of the function between x = -1 and x = 1. By using the techniques of factoring, we can find that the root is x=3, which is not between -1 and 1.
5. Given the function k(x) = x^3+3x^2+3x+1/(x+1). We know that k(0) = 1 and k(1) = 2. Using the IVT, we can prove that there are no roots of the function between x = 0 and x = 1. By using the techniques of factoring, we can find that the function is not defined for x=-1, which is not between 0 and 1.
6. Consider the function f(x) = x^3 - 5x^2 + 7x - 3. We know that f(1) = -1 and f(2) = 4. Using the IVT, we can prove that there exists at least one root of the function between x = 1 and x = 2. By using the techniques of factoring, we can find that the roots are x = 1, x = 2, and x = 3.
7. Consider the function g(x) = sin(x). We know that g(0) = 0 and g(π/2) = 1. Using the IVT, we can prove that there exists at least one root of the function between x = 0 and x = π/2. By using the techniques of graphing, we can find that the root is approximately x = π/4.

Consider the function h(x) = e^x. We know that h(0) = 1 and h(1) = e. Using the IVT, we can prove that there exists at least one root of the function between x = 0 and x = 1. By using the techniques of graphing, we can find that the root is approximately x = 0.36788.
Consider the function j(x) = x^2 - 4x + 3. We know that j(-1) = 2 and j(2) = 5. Using the IVT, we can prove that there exists at least one root of the function between x = -1 and x = 2. By using the techniques of factoring, we can find that the roots are x = 1 and x = 3.
Consider the function k(x) = x^2 + x - 6. We know that k(-2) = -8 and k(2) = 2. Using the IVT, we can prove that there exists at least one root of the function between x = -2 and x = 2. By using the techniques of factoring, we can find that the roots are x = -3 and x = 2.

Summary

In summary, the Intermediate Value Theorem (IVT) is a powerful tool that can be used to prove the existence of roots for a function and solve problems. It states that for any value c between the minimum and maximum values of a continuous function, there exists a point at which the function takes on the value c. By understanding and applying the IVT, we can gain a better understanding of the behavior of functions and find solutions to problems.