/*! 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 42 Find functions \(f\) and \(g\) s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find functions \(f\) and \(g\) such that \(h=g \circ f .\) (Note: The answer is not unique.) \(h(x)=\frac{1}{\sqrt{2 x+1}}+\sqrt{2 x+1}\)

Short Answer

Expert verified
We can choose the functions \(f(x) = 2x+1\) and \(g(x) = \frac{1}{\sqrt{x}} + \sqrt{x}\). When composed, \(h(x) = g(f(x)) = g(2x+1) = \frac{1}{\sqrt{2x+1}}+\sqrt{2x+1}\).

Step by step solution

01

Analyze the given function h(x)

Given the function \(h(x) = \frac{1}{\sqrt{2x+1}}+\sqrt{2x+1},\) we can recognize two different expressions: \(\frac{1}{\sqrt{2x+1}}\) and \(\sqrt{2x+1}\). We will use these expressions to find the appropriate functions \(f\) and \(g\).
02

Choose function f(x)

Let's start by choosing the function \(f(x)\). We can select \(f(x)\) to be the expression inside the square root of the given function \(h(x)\). So, \(f(x) = 2x+1\).
03

Choose function g(x)

Now, we need to find the function \(g(x)\). Observe the relationship between \(f(x)\) and \(h(x)\). We have \(\frac{1}{\sqrt{f(x)}} + \sqrt{f(x)}\). Thus, we can choose the function \(g(x)\) as follows: \(g(x) = \frac{1}{\sqrt{x}} + \sqrt{x}\).
04

Compose the functions g and f

Finally, let's compose the functions \(g\) and \(f\) to find \(h(x)\). \(h(x) = g(f(x)) = g(2x+1) = \frac{1}{\sqrt{2x+1}}+\sqrt{2x+1}\). Hence, we have found the functions \(f\) and \(g\) as \(f(x) = 2x+1\) and \(g(x) = \frac{1}{\sqrt{x}} + \sqrt{x}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Functions
Understanding composite functions is like learning to piece together a puzzle. These functions occur when you combine two functions in a way where the output of one becomes the input of the other. This combination creates a new function, often denoted as \( g \circ f \), which can be read as 'g composed with f'.

In the exercise provided, when we search for functions \( f \) and \( g \) that create a composite function \( h \), it's vital to visually and analytically dissect the given function \( h(x) \) to identify how one function is nested within the other. As the solution suggests, \( f(x) \) was chosen as \( 2x+1 \) because it's the inner part of \( h(x) \)—the 'inside' function. \( g(x) \) was identified as \( \frac{1}{\sqrt{x}} + \sqrt{x} \) because it properly transforms the output of \( f(x) \) into the final form of \( h(x) \). This interplay is at the heart of creating composite functions, where \( h(x) = g(f(x)) \).
Square Root Functions
Delving into square root functions, we uncover the essence of extracting roots and how they influence the shape and behavior of graphs. The basic form of a square root function is \( f(x) = \sqrt{x} \), representing a radical expression with an index of two.

In the context of the exercise, the function \( h(x) \) incorporates square roots in two contrasting manners, resulting in expressions \( \frac{1}{\sqrt{2x+1}} \) and \( \sqrt{2x+1} \). Handling these correctly is pivotal as it gives us the clues to formulate \( f(x) \) and \( g(x) \) for solving the composite function. When you encounter square root functions, pay attention to their domain—since you cannot take the square root of a negative number without entering the realm of complex numbers—and to their behavior, which is closely related to the concept of radical equations and their transformations.
Function Operations
Function operations, including addition, subtraction, multiplication, and division, extend our mathematical toolkit to alter and combine functions in various ways, much like basic arithmetic with numbers. These operations enable us to create complex functions from simple ones.

Within our exercise, function operations play a crucial role in decomposing the given \( h(x) \) into functions \( f \) and \( g \) that can be reassembled into its original form. This process involves recognizing that \( h(x) \) is the result of an addition operation between \( \frac{1}{\sqrt{2x+1}} \) and \( \sqrt{2x+1} \)—an example of function addition. In the broader scope, mastering function operations empowers you to manipulate and create complex expressions, paving the way for solving intricate equations and modeling real-world scenarios.
Applied Mathematics
Applied mathematics is the application of mathematical methods by different fields such as science, engineering, business, and industry. In this discipline, mathematical theories are translated into practical tools. Composite functions and operations on them are not just abstract concepts; they are widely used in various applications like calculating interest, predicting population growth, or even in computer algorithms.

The exercise above serves as a primer to seeing beyond numbers and symbols. When you come across problems like these, think about how you might apply similar methods to calculate a trajectory in physics or to optimize a function in economics. This level of understanding elevates your perspective from solving textbook exercises to appreciating the power and utility of mathematics in the real world.

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

Patricia's neighbor, Juanita, also wishes to have a rectangular-shaped garden in her backyard. But Juanita wants her garden to have an area of \(250 \mathrm{ft}^{2}\). Letting \(x\) denote the width of the garden, find a function \(f\) in the variable \(x\) giving the length of the fencing required to construct the garden. What is the domain of the function? Hint: Refer to the figure for Exercise 26. The amount of fencing required is equal to the perimeter of the rectangle, which is twice the width plus twice the length of the rectangle.

In Exercises 35-38, determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. A polynomial function is a sum of constant multiples of power functions.

PREVALENCE OF ALZHEIMER's PATIENTS Based on a study conducted in 1997 , the percent of the U.S. population by age afflicted with Alzheimer's disease is given by the function \(P(x)=0.0726 x^{2}+0.7902 x+4.9623 \quad(0 \leq x \leq 25)\) where \(x\) is measured in years, with \(x=0\) corresponding to age \(65 .\) What percent of the U.S. population at age 65 is expected to have Alzheimer's disease? At age 90 ?

The percentage of obese children aged \(12-19\) in the United States is approximately \(P(t)=\left\\{\begin{array}{ll}0.04 t+4.6 & \text { if } 0 \leq t<10 \\\ -0.01005 t^{2}+0.945 t-3.4 & \text { if } 10 \leq t \leq 30\end{array}\right.\) where \(t\) is measured in years, with \(t=0\) corresponding to the beginning of 1970 . What was the percentage of obese children aged \(12-19\) at the beginning of \(1970 ?\) At the beginning of 1985 ? At the beginning of 2000 ?

For years, automobile manufacturers had a monopoly on the replacement-parts market, particularly for sheet metal parts such as fenders, doors, and hoods, the parts most often damaged in a crash. Beginning in the late \(1970 \mathrm{~s}\), however, competition appeared on the scene. In a report conducted by an insurance company to study the effects of the competition, the price of an OEM (original equipment manufacturer) fender for a particular 1983 model car was found to be $$ f(t)=\frac{110}{\frac{1}{2} t+1} \quad(0 \leq t \leq 2) $$ where \(f(t)\) is measured in dollars and \(t\) is in years. Over the same period of time, the price of a non-OEM fender for the car was found to be $$ g(t)=26\left(\frac{1}{4} t^{2}-1\right)^{2}+52 \quad(0 \leq t \leq 2) $$ where \(g(t)\) is also measured in dollars. Find a function \(h(t)\) that gives the difference in price between an OEM fender and a non-OEM fender. Compute \(h(0), h(1)\), and \(h(2)\). What does the result of your computation seem to say about the price gap between OEM and non-OEM fenders over the 2 yr?

See all solutions

Recommended explanations on Math Textbooks

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.