/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Find \(f(g(x))\) and \(g(f(x))\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find \(f(g(x))\) and \(g(f(x))\) and determine whether each pair of functions \(f\) and \(g\) are inverses of each other. $$f(x)=5 x-9 \text { and } g(x)=\frac{x+5}{9}$$

Short Answer

Expert verified
Yes, the functions \(f(x) = 5x - 9\) and \(g(x) = \frac{x+5}{9}\) are inverses of each other as the compositions \(f(g(x))\) and \(g(f(x))\) both return \(x\).

Step by step solution

01

Calculate \(f(g(x))\)

Begin by plugging the function \(g(x) = \frac{x+5}{9}\) into the function \(f(x) = 5x - 9\). It would look like this: \(f(g(x)) = 5(\frac{x+5}{9}) - 9\). Simplify this to get \(f(g(x)) = x\).
02

Calculate \(g(f(x))\)

Now, you need to plug the function \(f(x) = 5x - 9\) into the function \(g(x) = \frac{x+5}{9}\). It would look like this: \(g(f(x)) = \frac{(5x - 9) + 5}{9}\). Simplify this to get \(g(f(x)) = x\).
03

Conclude if \(f\) and \(g\) are inverses

Since the results from Step 1 and Step 2 are both \(x\), it can be concluded that \(f\) and \(g\) are indeed inverses of each other because \(f(g(x)) = x\) and \(g(f(x)) = x\). This implies that for every \(x\) in the domain, the compositions of \(f\)and \(g\) still return \(x\) as the result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.