Problem 12 - Precalculus : Building Concepts and Connections

Question

In Exercises \(7-16,\) for the given functions \(f\) and \(g,\) find each composite function and identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \(\left(\frac{f}{g}\right)(x)\) $$f(x)=\frac{2}{x+1} ; g(x)=\frac{-1}{x^{2}}$$

Step-by-Step Solution

Answer

The composite functions and their domains are: (a) \((f+g)(x) =\frac{2x-1}{x^{2} + x}\) , Domain: All real numbers except -1. (b) \((f-g)(x) =\frac{2x^{2} +2x + 1}{x^{2} + x}\), Domain: All real numbers except -1. (c) \((fg)(x) =\frac{-2}{x^{3} + x^{2}}\), Domain: All real numbers except 0 and -1.(d) \((f/g)(x) = -2(x + 1)x^{2}\), Domain: All real numbers except 0.

Step 1: Identify the Functions

The functions provided in the question are:\(f(x)=\frac{2}{x+1}\) and \(g(x)=\frac{-1}{x^{2}}\)

Step 2: Calculate the composite function (f+g)(x)

(f+g)(x) is the addition of the two functions. So, \((f+g)(x) = f(x) + g(x) = \frac{2}{x+1} + \frac{-1}{x^{2}}\). After simplification, we get \((f+g)(x) =\frac{2x-1}{x^{2} + x}\) . The domain is all real numbers except -1, because for x=-1, the denominator becomes zero.

Step 3: Calculate the composite function (f-g)(x)

(f-g)(x) is the subtraction of the two functions. So, \((f-g)(x) = f(x) - g(x) = \frac{2}{x+1} - \frac{-1}{x^{2}}\). After simplification, we get \((f-g)(x) =\frac{2x^{2} +2x + 1}{x^{2} + x}\). The domain is all real numbers except -1 for similar reasons to step 2.

Step 4: Calculate the composite function (fg)(x)

(fg)(x) is the multiplication of the two functions. So, \((fg)(x) = f(x) * g(x) = \frac{2}{x+1} * \frac{-1}{x^{2}}\). After simplification, we get \((fg)(x) =\frac{-2}{x^{3} + x^{2}}\). The domain excludes 0 and -1, where the denominator goes to zero.

Step 5: Calculate the composite function (f/g)(x)

\((f/g)(x)\) is the division of the two functions. So, \((f/g)(x) = f(x) / g(x) = \frac{2}{x+1} / \frac{-1}{x^{2}}\) . After simplification, we get \((f/g)(x) = -2(x + 1)x^{2}\). The domain excludes 0, where the denominator would be zero.

Key Concepts

Function Composition

Understanding function composition is essential in precalculus, particularly when studying composite functions like \( (f+g)(x) \) or \( (f/g)(x) \). A composite function is created when one function is applied and then another function is applied to the result of the first. It's like a chain reaction: you start with an input, change it with the first function, and then the output of that becomes the input for the second function.

For instance, if \( f(x) = \frac{2}{x+1} \) and \( g(x) = \frac{-1}{x^{2}} \)s, when we form \( (f+g)(x) \) we are essentially adding the outputs of \( f(x) \) and \( g(x) \) for any input \( x \). This operation combines the functions into a new function that can be simplified to a single expression, such as \( (f+g)(x) = \frac{2x-1}{x^{2} + x} \). Quite intriguingly, this blend of functions creates a brand new function with its own distinct properties and domain.

Domain of a Function

The domain of a function is the set of all possible inputs (or 'x' values) that the function can accept without causing mathematical complications like division by zero. It is a critical concept because it tells us which values are 'safe' to use with the function and which aren't.

For example, in the composition \( (f/g)(x) \), where \( f(x) \) and \( g(x) \) are as previously specified, the domain would exclude values that make the denominator equal to zero. In this case, we cannot use \( x = 0 \) because it would make the denominator of \( g(x) \) zero, causing an undefined situation in the division. Hence, stating the domain is an essential part of working with functions, to avoid undefined or non-real results.

Rational Functions

Rational functions are ratios of two polynomials. They take the form \( \frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) \) is not the zero polynomial. They are particularly interesting because their graphs can have vertical asymptotes and holes, which arise from values that make \( q(x) \) equal zero; this ties back to the domain of the function.

In our textbook example, the functions \( f(x) \) and \( g(x) \) are both rational. When these rational functions are combined into composite functions such as \( (f-g)(x) \) or \( (fg)(x) \) or even \( (f/g)(x) \) we often deal with more complex rational functions. Simplifying these gives us reduced expressions that better reveal their domains and behaviors. Rational functions exemplify the beauty of algebra, where complex relationships give way to deeper understanding through simplification and analysis.