Chapter 6: Problem 4
If \(f(x)=\frac{x-3}{x+1}\) and \(g(x)=\frac{1}{x}\) find \(g(f(x))\).
Short Answer
Expert verified
Answer: The composition of the functions \(g(f(x))\) is \(\frac{x+1}{x-3}\).
Step by step solution
01
Identify the given functions
We are given two functions, \(f(x)=\frac{x-3}{x+1}\) and \(g(x)=\frac{1}{x}\).
02
Find the composition of the functions
To find the composition of the functions, we need to evaluate the function \(g\) at \(f(x)\). In other words, we need to replace \(x\) in the function \(g(x)\) with the expression for \(f(x)\):
$$
g(f(x)) = g\left(\frac{x-3}{x+1}\right)
$$
03
Evaluate the function \(g\) at \(f(x)\)
To evaluate the function \(g\) at \(f(x)\), replace \(x\) in the function \(g(x)\) with the expression for \(f(x)\):
$$
g\left(\frac{x-3}{x+1}\right) = \frac{1}{\frac{x-3}{x+1}}
$$
04
Simplify the expression
To simplify the expression, we can use the rule for the division of fractions:
$$
\frac{1}{\frac{x-3}{x+1}} = \frac{1}{1} \cdot \frac{x+1}{x-3} = \frac{x+1}{x-3}
$$
So, the composition of the functions, \(g(f(x))\), is \(\frac{x+1}{x-3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Composite Functions
Understanding composite functions is a vital part of algebra that involves combining two functions to create a third function. Computation of composite functions involves the insertion of one function into another.
To find the composite of two functions, denoted as \(g(f(x))\), we start by determining \(f(x)\), then replacing every \(x\) in \(g(x)\) with that expression. In our given exercise, \(f(x)=\frac{x-3}{x+1}\) and \(g(x)=\frac{1}{x}\), so when we create a composite, we essentially evaluate \(g\) at \(f(x)\).
Breaking down our example step by step, we find \(g\big(f(x)\big)\) by substituting \(\frac{x-3}{x+1}\) into \(g(x)\), yielding \(g\big(\frac{x-3}{x+1}\big) = \frac{1}{\frac{x-3}{x+1}}\), which we can then simplify. This process showcases the fundamental mechanics of creating and understanding composite functions.
To find the composite of two functions, denoted as \(g(f(x))\), we start by determining \(f(x)\), then replacing every \(x\) in \(g(x)\) with that expression. In our given exercise, \(f(x)=\frac{x-3}{x+1}\) and \(g(x)=\frac{1}{x}\), so when we create a composite, we essentially evaluate \(g\) at \(f(x)\).
Breaking down our example step by step, we find \(g\big(f(x)\big)\) by substituting \(\frac{x-3}{x+1}\) into \(g(x)\), yielding \(g\big(\frac{x-3}{x+1}\big) = \frac{1}{\frac{x-3}{x+1}}\), which we can then simplify. This process showcases the fundamental mechanics of creating and understanding composite functions.
Simplifying Expressions
Simplifying expressions is the process of breaking down complex algebraic expressions into their simplest form. This often involves applying algebraic rules, factoring, canceling terms, and reducing fractions.
In the context of our problem, simplifying begins once we've composed the functions. We're left with a complex fraction \(\frac{1}{\frac{x-3}{x+1}}\). To simplify, we use the reciprocal property of division, flipping the numerator and denominator of the inner fraction and multiplying. This results in a simpler fraction, \(\frac{x+1}{x-3}\), which is easier to evaluate and understand. Simplification is a key step to ensure that functions are presented in the most accessible form possible, allowing for clearer analysis and graphing.
In the context of our problem, simplifying begins once we've composed the functions. We're left with a complex fraction \(\frac{1}{\frac{x-3}{x+1}}\). To simplify, we use the reciprocal property of division, flipping the numerator and denominator of the inner fraction and multiplying. This results in a simpler fraction, \(\frac{x+1}{x-3}\), which is easier to evaluate and understand. Simplification is a key step to ensure that functions are presented in the most accessible form possible, allowing for clearer analysis and graphing.
Algebraic Functions
Algebraic functions are representations of relationships between variables through algebraic expressions. Examples include linear, quadratic, polynomial, and rational functions. In our exercise, we deal with rational functions, which are characterized by the division of two polynomials.
The function \(f(x)\) in the exercise is a rational function. It's important to handle such functions with care, particularly noting the domain restrictions—in this case, \(xeq -1\), because division by zero is undefined. Algebraic functions are the building blocks of higher mathematics and play a crucial part in modeling real-world scenarios. Understanding how to manipulate these functions, including operations like function composition, is foundational for advanced algebra.
The function \(f(x)\) in the exercise is a rational function. It's important to handle such functions with care, particularly noting the domain restrictions—in this case, \(xeq -1\), because division by zero is undefined. Algebraic functions are the building blocks of higher mathematics and play a crucial part in modeling real-world scenarios. Understanding how to manipulate these functions, including operations like function composition, is foundational for advanced algebra.
Inverse Functions
Inverse functions reverse the effect of the original function. They are a pair of functions that undoes each other. For a function \(f(x)\) with an inverse \(f^{-1}(x)\), applying \(f^{-1}(x)\) after \(f(x)\) will return to the starting value of \(x\). In symbolic terms, \(f^{-1}(f(x)) = x\).
Our exercise doesn't directly deal with inverse functions, but understanding the concept is useful for comprehensive knowledge in function manipulation. The process of finding an inverse function typically involves swapping the variables and solving for the new 'output' variable, which often requires simplifying algebraic expressions. It's essential to know that not all functions have inverses, especially if they are not one-to-one, which means that for every \(x\), there is a unique \(y\), and vice versa.
Our exercise doesn't directly deal with inverse functions, but understanding the concept is useful for comprehensive knowledge in function manipulation. The process of finding an inverse function typically involves swapping the variables and solving for the new 'output' variable, which often requires simplifying algebraic expressions. It's essential to know that not all functions have inverses, especially if they are not one-to-one, which means that for every \(x\), there is a unique \(y\), and vice versa.
