Understanding the Intermediate Value Theorem

The video tutorial explores the Intermediate Value Theorem, focusing on its application and limitations. It first examines whether the function g(x) = 1/x can be used with the theorem over the interval [-1, 1], concluding it cannot due to discontinuity at x = 0. The tutorial then successfully applies the theorem to the interval [1, 2], demonstrating that a solution exists for g(x) = 3/4 within this range.

What is the function g(x) defined as in the initial problem?

• g(x) = x

• g(x) = 1/x

• g(x) = x^2

• g(x) = x + 1

Why can't the Intermediate Value Theorem be applied to g(x) = 1/x over the interval [-1, 1]?

• The function is not continuous over the interval

• The function is not defined at x = -1

• The function is not defined at x = 1

• The function is continuous over the interval

What is the new interval considered for the function g(x) = 1/x?

• [-1, 1]

• [0, 1]

• [1, 2]

• [2, 3]

What must be true about a function for the Intermediate Value Theorem to be applicable?

• It must be defined at all points

• It must be continuous over the interval

• It must be differentiable

• It must be increasing

What are the values of g(x) at the endpoints of the interval [1, 2]?

• g(1) = 1, g(2) = 1/2

• g(1) = 1/2, g(2) = 1

• g(1) = 1, g(2) = 2

• g(1) = 2, g(2) = 1

What value is between g(1) and g(2) in the interval [1, 2]?

• 1/4

• 1

• 1/2

• 3/4

What does the Intermediate Value Theorem guarantee in the interval [1, 2] for g(x) = 1/x?

• There is no solution

• There is a solution for g(x) = 3/4

• The function is not continuous

• The function is undefined

What is the main requirement for applying the Intermediate Value Theorem?

• The function must be constant

• The function must be decreasing

• The function must be continuous over the interval

• The function must be increasing

What is the conclusion about the function g(x) = 1/x over the interval [1, 2]?

• The function is not continuous

• The function is decreasing

• The function is continuous and the theorem applies

• The function is undefined

What does the Intermediate Value Theorem help to find in this context?

• The derivative of the function

• The maximum value of the function

• The minimum value of the function

• A solution for g(x) = 3/4