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Understanding Functions and Their Inverses

Understanding Functions and Their Inverses

Assessment

Interactive Video

Mathematics

9th - 12th Grade

Practice Problem

Hard

CCSS
HSF-BF.B.4A, HSF-BF.B.4C

Standards-aligned

Created by

Aiden Montgomery

FREE Resource

Standards-aligned

CCSS.HSF-BF.B.4A
,
CCSS.HSF-BF.B.4C
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.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

x ≥ 1

x ≥ 0

x ≥ 1/2

x > 0

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

[1/2, ∞)

[1, ∞)

(1/2, ∞)

[0, ∞)

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

y ≥ 0

y ≥ -3

y > -3

y > 0

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

[0, ∞)

(-3, ∞)

(-∞, -3]

[-3, ∞)

Tags

CCSS.HSF-BF.B.4A

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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.

Tags

CCSS.HSF-BF.B.4A

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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.

Tags

CCSS.HSF-BF.B.4A

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

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