Domain and Range of Functions

Domain and Range of Functions

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Hard

Created by

Thomas White

FREE Resource

Anil Kumar explains how to find the inverse of a square root function. The video begins with an introduction to the problem and proceeds to analyze the domain and range of the given function. The process of finding the inverse function involves swapping X and Y and solving for Y. The video also addresses corrections to the domain and range for the inverse function, ensuring clarity in the explanation. The conclusion includes clarifications and corrections to any mistakes made during the explanation.

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16 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main topic discussed in the video?

Finding the inverse of square root functions

Solving quadratic equations

Graphing linear functions

Understanding logarithmic functions

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

X is less than or equal to 1

X is less than or equal to 0

X is greater than or equal to 1

X is greater than or equal to 0

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

Y is greater than or equal to -3

Y is greater than or equal to 3

Y is greater than or equal to 0

Y is greater than or equal to 1

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

Solve for X

Graph the function

Swap X and Y

Check the domain

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens to the domain and range when finding the inverse?

They are halved

They are doubled

They are swapped

They remain the same

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

After swapping X and Y, what is the next step?

Solve for Y

Graph the inverse

Check the range

Swap X and Y again

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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

y = 4(x-3)^2 - 1

y = 4(x+3)^2 + 1

y = 4(x-3)^2 + 1

y = 4(x+3)^2 - 1

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