Understanding Functions and Their Inverses

The video tutorial explains how to determine the domain and range of a function involving a square root, and how to find its inverse. It covers the restrictions on the domain due to the square root, and how to express the domain and range using interval notation. The tutorial also demonstrates how to find the inverse function by interchanging variables and solving for y, and verifies the results graphically by showing symmetry across the line y = x.

What is the domain of the function f(x) = √(2x - 1) - 3?

• x ≥ 1

• x ≥ 0

• x ≥ 1/2

• x > 0

In interval notation, how is the domain of f(x) = √(2x - 1) - 3 expressed?

• [1/2, ∞)

• [1, ∞)

• (1/2, ∞)

• [0, ∞)

What is the range of the function f(x) = √(2x - 1) - 3?

• y ≥ 0

• y ≥ -3

• y > -3

• y > 0

How is the range of f(x) = √(2x - 1) - 3 expressed in interval notation?

• [0, ∞)

• (-3, ∞)

• (-∞, -3]

• [-3, ∞)

What happens to the domain of a function when finding its inverse?

• It becomes the domain of the inverse function.

• It remains the same.

• It becomes the range of the inverse function.

• It is not related to the inverse function.

What is the first step in finding the inverse of a function?

• Solve for x in terms of y.

• Interchange x and y.

• Graph the function.

• Find the domain of the function.

What is the inverse function of f(x) = √(2x - 1) - 3?

• f⁻¹(x) = 2(x + 3)² + 1

• f⁻¹(x) = 1/2(x + 3)² + 1

• f⁻¹(x) = (x + 3)² - 1

• f⁻¹(x) = 1/2(x - 3)² + 1

What is the domain of the inverse function f⁻¹(x) = 1/2(x + 3)² + 1?

• x ≥ -3

• x > 1/2

• x ≥ 1/2

• x > -3

How can you verify graphically that two functions are inverses of each other?

• They intersect at the origin.

• They are symmetrical across the line y = x.

• They have the same domain.

• They have the same range.

What does the graph of y = x represent in the context of inverse functions?

• The x-axis.

• The range of the original function.

• The domain of the inverse function.

• The line of symmetry for the function and its inverse.