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In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=4 x-1 ; g(x)=\frac{x+1}{4}$$

Short Answer

Expert verified
The composition functions \(f(g(x))\) and \(g(f(x))\) result in \(x\) for both, and their domains are all real numbers as expressed by the interval \(-\infty, \infty\).

Step by step solution

01

Find the function composition \(f \circ g\)

To find the function \(f \circ g\), replace all occurrences of \(x\) in \(f(x)\) with \(g(x)\). So, \(f(g(x))=4\left(\frac{x+1}{4}\right)-1\). Simplify this expression to get \(f(g(x))=x+1-1=x\)
02

Find the domain of the function composition \(f \circ g\)

The domain of \(f \circ g\) is all values of \(x\) for which \(g(x)\) is defined and the resulting value \(g(x)\) is in the domain of \(f\). Since \(g(x)=\frac{x+1}{4}\) is defined for all real numbers and all real numbers are in the domain of \(f(x)=4 x-1\), the domain of \(f \circ g\) is \(-\infty, \infty\).
03

Find the function composition \(g \circ f\)

To find the function \(g \circ f\), replace all occurrences of \(x\) in \(g(x)\) with \(f(x)\). So, \(g(f(x))=\frac{4x-1+1}{4}\). Simplify this expression to get \(g(f(x))=x\).
04

Find the domain of the function composition \(g \circ f\)

The domain of \(g \circ f\) is all values of \(x\) for which \(f(x)\) is defined and the resulting value \(f(x)\) is in the domain of \(g\). Since \(f(x)=4 x-1\) is defined for all real numbers and all real numbers are in the domain of \(g(x)=\frac{x+1}{4}\), so the domain of \(g \circ f\) is \(-\infty, \infty\).

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Key Concepts

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

Domain of a Function
The domain of a function represents the set of all possible input values for which the function is defined. It's essential to determine the domain to understand the limitations and behavior of the function. For example, if a function includes a denominator, we must exclude any values that would make the denominator zero, as division by zero is undefined.

When analyzing the domain in a function composition, such as \(f \circ g\), we must consider both the domain of \(g\) and whether the outputs of \(g\) fall within the domain of \(f\). If \(g(x) = \frac{x+1}{4}\), the function is defined for all real numbers because there is no restriction (like division by zero or square roots of negative numbers). The composite function \(f \circ g\) then maintains this unlimited domain, as \(f\)'s operations do not impose new restrictions.
Composite Functions
A composite function is formed when one function is applied to the result of another function. Notation such as \(f \circ g\) represents the function \(f\) composed with \(g\), which means you apply \(g\) first and then \(f\) to the result. The expression for \(f \circ g(x)\) is obtained by substituting \(g(x)\) into \(f\). In our example, applying \(g\) to \(x\) gives us \(g(x) = \frac{x+1}{4}\), and then applying \(f\) to \(g(x)\) results in \(f(g(x)) = x\), which shows a sort of 'cancelling out' effect, simplifying back to \(x\).

This process can sometimes lead to surprising simplifications, where complex expressions reduce down to much simpler forms. It's a powerful tool in precalculus that helps us do things like solve equations involving function compositions or analyze their properties.
Precalculus
The field of precalculus serves as the bridge between the concrete world of algebra and the more abstract realm of calculus. It encompasses a variety of topics, including function composition and the concept of domains. Precalculus equips students with the necessary skills to tackle calculus, such as understanding how different functions behave and interact, and how to manipulate algebraic expressions to solve problems.

The exercise involving \(f\) and \(g\) is a typical precalculus problem. It challenges students to think about how functions can be combined and requires understanding of how to navigate different operations within functions. As such, students engaged in precalculus coursework will frequently encounter these kinds of exercises, which are designed to enhance their ability to think critically about functions before moving on to the study of calculus.

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Most popular questions from this chapter

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(-2 x+5)^{2}$$

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A rectangular garden plot is to be enclosed with a fence on three of its sides and an existing wall on the fourth side. There is 45 feet of fencing material available. (a) Write an equation relating the amount of available fencing material to the lengths of the three sides that are to be fenced. (b) Use the equation in part (a) to write an expression for the width of the enclosed region in terms of its length. (c) For each value of the length given in the following table of possible dimensions for the garden plot, fill in the value of the corresponding width. Use your expression from part (b) and compute the resulting area. What do you observe about the area of the enclosed region as the dimensions of the garden plot are varied? $$\begin{array}{ccc}\hline \begin{array}{c}\text { Length } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Width } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Total Amount of } \\\\\text { Fencing Material (feet) }\end{array} & \begin{array}{c}\text { Area } \\\\\text { (square feet) }\end{array} \\\\\hline 5 & & 45 \\\10 & & 45 \\\15 & & 45 \\\20 & & 45 \\\30 & & 45 \\\k & & 45 \\\& &\end{array}$$ (d) Write an expression for the area of the garden plot in terms of its length. (e) Find the dimensions that will yield a garden plot with an area of 145 square feet.

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=\sqrt[3]{4 x^{2}-1}$$

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