Question

For each function, a) determine whether it is one-to-one; b) if it is one-to-one, determine its inverse function. $k(x)=2 x-7$

Name: Problem 52, Chapter 9: Intermediate Algebra for College Students
Uploaded: 2021-09-08T13:43:07-08:00
Duration: 56 s
Channel: Yujie Wang
Description: For each function, a) determine whether it is one-to-one; b) if it is one-to-one, determine its inverse function. k(x)=2 x-7

   For each function, a) determine whether it is one-to-one; b) if it is one-to-one, determine its inverse function.
$k(x)=2 x-7$

Intermediate Algebra for College Students

Allen R. Angel;… 10th Edition

Chapter 9, Problem 52

Instant Answer

Solved by Expert Yujie Wang

09/08/2021

Step 1

In other words, for any two different inputs x1 and x2, their outputs k(x1) and k(x2) should be different. Let's assume k(x1) = k(x2) for some x1 and x2. Then we have: 2x1 - 7 = 2x2 - 7 2x1 = 2x2 x1 = x2 Since x1 = x2, this means that the function is Show more…

Show all steps

Thanks for your feedback!

For each function, a) determine whether it is one-to-one; b) if it is one-to-one, determine its inverse function. $k(x)=2 x-7$

Play audio

Feedback

Yujie Wang and 94 other educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

One-to-One Function

A function is one-to-one (injective) if no two different input values produce the same output. This characteristic is crucial because it ensures that each output is uniquely associated with a single input, making it possible to reverse the function and retrieve the original input from a given output.

Inverse Function

An inverse function undoes the action of the original function. It is defined by swapping the roles of the input and output variables and then solving for the new output. A function must be one-to-one for its inverse to exist, as this guarantees that every element in the range corresponds to a unique element in the domain.

Please give Ace some feedback