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Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$f(x)=\frac{x+2}{x-1}, \quad g(x)=\frac{x-5}{x+4}$$

Short Answer

Expert verified
(a) \( (f \circ g)(x) = \frac{x+3}{x-9} \), domain: all \(x \neq -4, 9\). (b) \((g \circ f)(x) = \frac{-4x+7}{5x-2}\), domain: all \(x \neq 1, \frac{2}{5}\).

Step by step solution

01

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

The composite function \((f \circ g)(x)\) is defined as \(f(g(x))\). Substitute \(g(x) = \frac{x-5}{x+4}\) into \(f(x)\): \[f(g(x)) = \frac{\left(\frac{x-5}{x+4}\right) + 2}{\left(\frac{x-5}{x+4}\right) - 1} = \frac{\frac{x-5}{x+4} + \frac{8}{4}}{\frac{x-5}{x+4} - \frac{4}{4}}\] Simplify each part of the fraction: \[f(g(x)) = \frac{\frac{x-5 + 8}{x+4}}{\frac{x-5-4}{x+4}} = \frac{\frac{x+3}{x+4}}{\frac{x-9}{x+4}}\] By multiplying by the reciprocal of the denominator, we obtain: \[f(g(x)) = \frac{x+3}{x-9}\]
02

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

The domain of \(f \circ g\) consists of all \(x\) values that are within the domains of both \(g(x)\) and \(f(g(x))\). 1. For \(g(x)\), the domain is all real numbers except \(x = -4\) where the denominator is zero. 2. For \(f(g(x))\), set the denominator \(x-9\) not equal to zero: \(x eq 9\). Thus, the domain of \(f \circ g\) is all real numbers except \(x = -4\) and \(x = 9\).
03

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

The composite function \((g \circ f)(x)\) is defined as \(g(f(x))\). Substitute \(f(x) = \frac{x+2}{x-1}\) into \(g(x)\): \[g(f(x)) = \frac{\left(\frac{x+2}{x-1}\right) - 5}{\left(\frac{x+2}{x-1}\right) + 4} = \frac{\frac{x+2-5(x-1)}{x-1}}{\frac{x+2+4(x-1)}{x-1}}\] Simplify each part of the fraction: \[g(f(x)) = \frac{\frac{x+2-5x+5}{x-1}}{\frac{x+2+4x-4}{x-1}} = \frac{\frac{-4x+7}{x-1}}{\frac{5x-2}{x-1}}\] By multiplying by the reciprocal of the denominator, we obtain: \[g(f(x)) = \frac{-4x+7}{5x-2}\]
04

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

The domain of \(g \circ f\) consists of all \(x\) values that are within the domains of both \(f(x)\) and \(g(f(x))\). 1. For \(f(x)\), the domain is all real numbers except \(x = 1\) where the denominator is zero. 2. For \(g(f(x))\), set the denominator \(5x-2\) not equal to zero: \(x eq \frac{2}{5}\). Thus, the domain of \(g \circ f\) is all real numbers except \(x = 1\) and \(x = \frac{2}{5}\).

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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 refers to the set of all possible input values (usually, the variable \(x\)) that won't cause any mathematical contradiction such as division by zero or taking the square root of a negative number in real numbers. It matters because it sets limits on the application of functions in various problems.

When dealing with simple functions like linear functions, the domain is typically all real numbers. However, for functions like rational functions, which involve fractions, the domain can be restricted. This is because dividing by zero is undefined in mathematics.

For instance, if you have a function \(f(x) = \frac{1}{x-1}\), the domain would exclude \(x = 1\) because it makes the denominator zero. Thus, the domain is all real numbers except \(x = 1\). Similarly, identifying the domain for composite functions requires ensuring both functions' domains do not cause undefined expressions.
Function Composition
Function composition is the process by which two functions, say \(f(x)\) and \(g(x)\), are combined into a single function. This is done by substituting one function into the other. The notation \(f(g(x))\) implies that you are inputting function \(g(x)\) into function \(f(x)\). When evaluating a composite function, the order of operations is crucial and sometimes can affect the result significantly.

In general, to perform function composition, you simply replace the variable in the outer function \(f(x)\) with the entire inner function \(g(x)\). Each step requires careful substitution and simplification. This process helps solve more complex problems where simple direct evaluation might not be possible due to the multi-step nature that composite functions introduce.

The challenge arises when determining the domain, as it is necessary to consider the restrictions of both functions involved. One must evaluate which values of \(x\) make either the inner or outer function undefined and exclude these from the domain.
Rational Functions
Rational functions are a type of function characterized by the ratio of two polynomials. They are represented in the form \(f(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) \(eq\) 0. Identifying the domain of a rational function involves finding the values where \(Q(x) = 0\), since division by zero is mathematically undefined.

Such functions can exhibit a variety of behaviors, like vertical asymptotes or discontinuities at points where the denominator equals zero and goes undefined. These characteristics are essential for understanding the behavior of the graph of the function.

To compute with rational functions, especially in composite function scenarios, it helps to simplify expressions by finding common terms that can be canceled. However, always keep in mind the excluded values which might "cancel" numerically but still are exclusions in the domain of the function, maintaining mathematical integrity.

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