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Chapter 2: Problem 71

Given functions \(f\) and \(g,\) find ( \(a\) ) \((f \circ g)(x)\) and its domain, and ( \(b\) ) \((g \circ f)(x)\) and its domain. See Examples 6 and 7 . $$f(x)=\frac{2}{x}, \quad g(x)=x+1$$

Short Answer

Expert verified
a) \((f \circ g)(x) = \frac{2}{x+1}\), domain: \( x eq -1 \). b) \((g \circ f)(x) = \frac{2}{x} + 1\), domain: \( x eq 0 \).

Step by step solution

01

- Understand Composite Functions

We are given two functions, \( f(x) = \frac{2}{x} \) and \( g(x) = x+1 \). Composite functions are formed by applying one function to the results of another. For \((f \circ g)(x)\), this means the function \( g(x) \) is applied first, and then \( f(x) \) is applied to the result. For \((g \circ f)(x)\), it means \( f(x) \) is applied first, and then \( g(x) \) is applied to the result.
02

- Find \((f \circ g)(x)\)

To find \((f \circ g)(x)\): 1. Substitute \( g(x) \) into \( f(x) \).2. \( g(x) = x + 1 \), so \( f(g(x)) = f(x + 1) \).3. Since \( f(x) = \frac{2}{x} \), substitute \( x + 1 \) for \( x \): \( f(x + 1) = \frac{2}{x + 1} \).Thus, \((f \circ g)(x) = \frac{2}{x+1} \).
03

- Determine the Domain of \((f \circ g)(x)\)

The domain of \( g(x) \) is all real numbers since \( g(x) = x + 1 \) is defined for all \( x \). However, the domain of \( f(x) \) is all real numbers except \( x = 0 \). So, the input to \( f \) (which is \( g(x) = x + 1 \)) must not be zero. This means \( x + 1 eq 0 \) or \( x eq -1 \). Therefore, the domain of \((f \circ g)(x)\) is all real numbers \( x eq -1 \).
04

- Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\): 1. Substitute \( f(x) \) into \( g(x) \).2. \( f(x) = \frac{2}{x} \), so \( g(f(x)) = g(\frac{2}{x}) \).3. Since \( g(x) = x + 1 \), substitute \( \frac{2}{x} \) for \( x \): \( g(\frac{2}{x}) = \frac{2}{x} + 1 \).Thus, \((g \circ f)(x) = \frac{2}{x} + 1 \).
05

- Determine the Domain of \((g \circ f)(x)\)

The domain of \( f(x) \) is all real numbers except \( x = 0 \) because \( f(x) = \frac{2}{x} \) is undefined for \( x = 0 \). Since \( \frac{2}{x} \) must be defined before applying \( g \), the domain of \((g \circ f)(x)\) is all real numbers \( x eq 0 \).

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Key Concepts

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

domain of functions
The domain of a function is the set of all possible input values (usually represented by the variable \(x\)) for which the function is defined. To find the domain, we need to identify the values of \(x\) that do not cause the function to produce undefined results, such as division by zero or taking the square root of a negative number.

For the functions given in the exercise:
  • \(f(x) = \frac{2}{x}\) has a domain of all real numbers except \(x = 0\) since division by zero is undefined.
  • \(g(x) = x + 1\) has a domain of all real numbers because there are no restrictions on adding 1 to any real number.

When dealing with composite functions like \((f \, \text{circ} \, g)(x)\) and \((g \, \text{circ} \, f)(x)\), it's important to consider the domains of both functions involved. Each step must produce a result that is within the domain of the next function.
function composition
Function composition involves applying one function to the result of another function. In notation, if we have two functions \(f(x)\) and \(g(x)\), the composite function \((f \, \text{circ} \, g)(x)\) means \(f(g(x))\), and \((g \, \text{circ} \, f)(x)\) means \(g(f(x))\).

To compose functions:
  • First, apply the inner function to the input \(x\).
  • Then, apply the outer function to the result of the first step.

In the given exercise:
  • For \((f \, \text{circ} \, g)(x)\): We apply \(g(x)\) first, which gives \(g(x) = x + 1\). Then, we use this result as the input for \(f\), i.e., \(f(x+1) = \frac{2}{x+1}\).
  • For \((g \, \text{circ} \, f)(x)\): We apply \(f(x)\) first, which gives \(f(x) = \frac{2}{x}\). Then, we use this result as the input for \(g\), i.e., \(g\left(\frac{2}{x}\right) = \frac{2}{x} + 1\).

Function composition requires careful consideration of the order in which functions are applied.
rational functions
A rational function is any function that can be expressed as the quotient of two polynomials. In mathematical terms, a rational function has the form \(R(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).

Rational functions often have restrictions on their domain because the denominator cannot be zero. This is why it's crucial to identify the values of \(x\) that lead to a zero denominator and exclude them from the domain.

In our exercise:
  • The function \(f(x) = \frac{2}{x}\) is a rational function. Since the denominator \(x\) cannot be zero, the domain is all real numbers except \(x = 0\).
  • The function \(f(g(x)) = \frac{2}{x+1}\) is also a rational function. The denominator \(x+1\) cannot be zero, so the domain is all real numbers except \(x = -1\).

Identifying the domain of composite rational functions, like \(f(g(x))\) and \(g(f(x))\), involves ensuring that the inputs do not create zero denominators at any step.

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Most popular questions from this chapter

Solve each problem. Emission of Pollutants When a thermal inversion layer is over a city (as happens in Los Angeles), pollutants cannot rise vertically but are trapped below the layer and must disperse horizontally. Assume that a factory smokestack begins emitting a pollutant at 8 A.M. Assume that the pollutant disperses horizontally over a circular area. If \(t\) represents the time, in hours, since the factory began emitting pollutants \((t=0 \text { represents } 8 \text { A.M.), assume that the radius of the circle of pollutants at time } t\) is \(r(t)=2 t\) miles. Let \(\mathscr{A}(r)=\pi r^{2}\) represent the area of a circle of radius \(r\) (a) Find \((\mathscr{A} \circ r)(t)\) (b) Interpret \((\mathscr{A} \circ r)(t)\) (c) What is the area of the circular region covered by the layer at noon?

Solve each problem. To visualize the situation, use graph paper and a pair of compasses to carefully draw the graphs of the circles. The locations of three receiving stations and the distances to the epicenter of an earthquake are contained in the following three equations: \((x-2)^{2}+(y-1)^{2}=25\) \((x+2)^{2}+(y-2)^{2}=16,\) and \((x-1)^{2}+(y+2)^{2}=9 .\) Determine the location of the epicenter.

Find the center-radius form of the circle described or graphed. a circle having a diameter with endpoints \((5,4)\) and \((-3,-2)\)

Solve each problem. To visualize the situation, use graph paper and a pair of compasses to carefully draw the graphs of the circles. Suppose that receiving stations \(P, Q,\) and \(R\) are located on a coordinate plane at the points \((3,1),(5,-4),\) and \((-1,4),\) respectively. The epicenter of an earthquake is determined to be \(\sqrt{5}\) units from \(P, 6\) units from \(Q,\) and \(2 \sqrt{10}\) units from \(R .\) Where on the coordinate plane is the epicenter located?

Find the function \(g(x)=a x+b\) whose graph can be obtained by translating the graph of \(f(x)=2 x+5\) up 2 units and to the left 3 units.
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