/*! 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 66 Suppose \(g\) is an even functio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 66

Suppose \(g\) is an even function and \(f\) is any function. Show that \(f \circ g\) is an even function.

Short Answer

Expert verified
The composition of an even function \(g\) and any function \(f\) results in an even function \(f \circ g\). To show this, we follow these steps: 1. Define the composition of the functions as \(h(x) = (f \circ g)(x)\), where \((f \circ g)(x) = f(g(x))\). 2. Express the composition at \(-x\), \((f \circ g)(-x) = f(g(-x))\). 3. Apply the evenness condition for the function \(g\), \(f(g(-x)) = f(g(x))\). 4. Conclude that \(f \circ g\) is an even function since \((f \circ g)(x) = (f \circ g)(-x)\) for all x in the domain of \(f \circ g\).

Step by step solution

01

Define the composition of the functions \(f\) and \(g\)

First, let's formally define the composition of the functions \(f\) and \(g\). This is given by the function \(h(x) = (f \circ g)(x)\), where \((f \circ g)(x) = f(g(x))\).
02

Express the composition function at \(-x\)

We need to find an expression for the composition at \(-x\). To do this, we substitute \(-x\) into our composition function, \((f \circ g)(-x)\), to get: \((f \circ g)(-x) = f(g(-x))\)
03

Apply the evenness condition for the function \(g\)

We know that \(g\) is an even function, which means that \(g(x) = g(-x)\) for all x in its domain. Applying this condition to the expression for \((f \circ g)(-x)\), we get: \(f(g(-x)) = f(g(x))\)
04

Conclude that \(f \circ g\) is an even function

Since we found that f(g(-x)) = f(g(x)) for all x in the domain of f(g(x)), we can see that our composition function, \(f \circ g\), satisfies the condition for an even function: \((f \circ g)(x) = (f \circ g)(-x)\) for all x in the domain of \(f \circ g\). Therefore, we can conclude that the composition of an even function \(g\) and any function \(f\) results in an even function \(f \circ g\).

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 91影视!

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Show that if \(f\) is the function defined by \(f(x)=m x+b,\) where \(m \neq 0,\) then \(f\) is a one-toone function.

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ What is the range of \(f ?\)

Find functions \(f\) and \(g\), each simpler than the given function \(h\), such that \(h=f \circ g\). \( h(x)=\frac{2}{3+\sqrt{1+x}}\)

Show that the composition of two increasing functions is increasing.

Check your answer by evaluating the appropriate function at your answer. Suppose \(f(x)=3 x+2\). Find a formula for \(f^{-1}\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.