Problem 65 - Precalculus Essentials

Question

Find. a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=\frac{1}{x}, g(x)=\frac{1}{x}$$

Step-by-Step Solution

Answer

The answers to the exercise are: a. \(f \circ g(x) = x\), b. \(g \circ f(x) = x\), c. \(f \circ g(2) = 2\), d. \(g \circ f(2) = 2\)

Step 1: Function Composition for \(f \circ g\)(x)

The function \(f \circ g(x)\) means that we are substituting \(g(x)\) into function \(f(x)\). Since \(g(x) = \frac{1}{x}\), we substitute this into \(f(x)\) to get: \(f(g(x)) = f\left(\frac{1}{x}\right)\) = \(\frac{1}{\frac{1}{x}} = x\)

Step 2: Function Composition for \(g \circ f\)(x)

The function \(g \circ f(x)\) means that we are substituting \(f(x)\) into function \(g(x)\). Since \(f(x) = \frac{1}{x}\), we substitute this into \(g(x)\) to get: \(g(f(x)) = g\left(\frac{1}{x}\right)\)= \(\frac{1}{\frac{1}{x}} = x\)

Step 3: Evaluate \(f \circ g\)(2)

To evaluate \(f \circ g(2)\), we use the function from step 1 where we found \(f \circ g(x) = x\). Substituting 2 into this equation, we get \(f \circ g(2) = 2\)

Step 4: Evaluate \(g \circ f\)(2)

To evaluate \(g \circ f(2)\), we use the function from step 2 where we found \(g \circ f(x) = x\). Substituting 2 into this equation, we get \(g \circ f(2) = 2\)

Key Concepts

Precalculus

Precalculus is a foundational mathematical course that prepares students for calculus and other higher-level math topics. It encompasses a variety of concepts, including functions, equations, and the study of changes in rates. Understanding precalculus is essential for students, as it sets the groundwork for the advanced mathematical reasoning required in calculus.

One fundamental aspect of precalculus involves the exploration of functions, which are relationships between a set of inputs and outputs. Students learn how to represent functions graphically, numerically, algebraically, and verbally. The skills gained in precalculus, such as manipulating algebraic expressions and understanding function properties, are vital for mastering function evaluation and function composition, both of which are also pillars of calculus.

Function Evaluation

Function evaluation is a basic yet integral part of working with mathematical functions. It involves finding the output of a function for a particular input. This concept comes in handy when analyzing graphs and solving problems in both precalculus and calculus.

For instance, given a function like f(x) = 1/x, evaluating the function for a value like x = 2 would mean substituting 2 into the function and calculating the result: f(2) = 1/2. By evaluating functions at specific points, students gain a better understanding of the behavior of functions and can plot them effectively on a graph.

Composite Functions

Composite functions, denoted as (f \(circ\) g)(x), involve combining two functions where the output of one function becomes the input to another. In essence, by creating composite functions, we can build new functions from existing ones.

To compute a composite function, one must first evaluate the inner function and then use that result as the input for the outer function. As shown in the textbook solution, for f(x) = 1/x and g(x) = 1/x, the composite function (f \(circ\) g)(x) results in simply x. This reveals that even with sophisticated operations like function composition, the outcomes can still be fundamentally simple and elegant.

Mathematical Functions

Mathematical functions are fundamental in various fields of math, representing relationships between two quantities where each input is related to exactly one output. They are like machines where every item that goes in comes out transformed in a certain way.

For example, with the functions f(x) = 1/x and g(x) = 1/x, we see that each function has its distinct rule of transformation. Functions can be utilized for diverse purposes, such as modeling real-world phenomena or expressing mathematical relationships. Proper comprehension of how to handle different types of functions allows students to solve a broad spectrum of mathematical problems efficiently and confidently.