Problem 58 - College Algebra

Question

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$ f(x)-x^{2}+1, g(x)=x^{2}-3 $$

Step-by-Step Solution

Answer

a. \( (f \circ g)(x) = x^{4} - 6x^{2} + 10 \) b. \( (g \circ f)(x) = x^{4} + 2x^{2} - 2 \) c. \( (f \circ g)(2) = 4 \) d. \( (g \circ f)(2) = 10 \)

Step 1: Find \( (f \circ g)(x) \)

First, to find \( (f \circ g)(x) \), you plug \( g(x) \) into \( f(x) \). In this case, \( g(x) = x^{2} - 3 \) and \( f(x) = x^{2} + 1 \). So, \( (f \circ g)(x) = (g(x))^{2} + 1 = (x^{2} - 3)^{2} + 1 = x^{4} - 6x^{2} + 10..

Step 2: Find \( (g \circ f)(x) \)

Next, to find \( (g \circ f)(x) \), you plug \( f(x) \) into \( g(x) \). In this case, \( f(x) = x^{2} + 1 \) and \( g(x) = x^{2} - 3 \). So, \( (g \circ f)(x) = (f(x))^{2} - 3 = (x^{2} + 1)^{2} - 3 = x^{4} + 2x^{2} - 2.

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

Now, to find \( (f \circ g)(2) \), you plug 2 into your solution for \( (f \circ g)(x) \) from Step 1. Thus, \( (f \circ g)(2) = (2^{4} - 6*2^{2} + 10) = 4.

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

Lastly, to find \( (g \circ f)(2) \), you plug 2 into your solution for \( (g \circ f)(x) \) from Step 2. Thus, \( (g \circ f)(2) = (2^{4} + 2*2^{2} - 2) = 10.

Key Concepts

Function Operations

Function operations are the ways in which we can combine, manipulate, and create new functions from existing ones. Understanding how to work with functions in algebra is essential because it sets the foundation for studying more complicated mathematical concepts. There are several basic operations for functions such as addition, subtraction, multiplication, division, and composition.

When we add or subtract functions, we simply add or subtract their outputs for any input x. Similarly, for multiplication and division, we multiply or divide their outputs. However, when we talk about the composition of functions, which is a more involved operation, we're referring to taking the output of one function and using it as the input for another function. This operation is denoted by \f \( \circ \) g\(x\) and creates an entirely new function with its own properties and behavior.

Composition

Understanding Composition of Functions

Composition of functions is a bit like a relay race. In a relay, one runner hands off a baton to the next runner. Similarly, with composition, the output of the first function \(g(x)\) becomes the input for the second function \(f(x)\). The notation we use for this is \(f \circ g\)(x), which reads as 'f composed with g at x'.

When we want to find \(f \circ g\)(x), we take \(g\)'s output and put it into \(f\). This might look complicated at first, but once you understand that it's all about performing \(f\)'s operation on \(g\)'s result, it becomes much clearer. It is important to note that the composition is not commutative - this means \(f \circ g\) is generally not equal to \(g \circ f\).

Algebraic Functions

Algebraic functions are functions that can be constructed using algebraic operations—addition, subtraction, multiplication, division, and taking roots—on polynomial functions. Examples include linear functions, quadratic functions, and rational functions.

An understanding of the basic shapes and properties of these algebraic functions helps in predicting the behavior of more complex functions. When we talk about the composition of functions, we are essentially creating a new algebraic function from existing ones, based on these operations. For example, when we substitute \(g(x) = x^2 - 3\) into \(f(x) = x^2 + 1\), we use function operations to create a new algebraic function which combines their properties.

Function Evaluation

What Does It Mean to Evaluate a Function?

Evaluating a function means finding the output of a function for a particular input value. This process often involves substituting a given number or expression into the function in place of the variable and simplifying the result.

It is straightforward when dealing with basic functions, but when it comes to composed functions like \(f \circ g\)(x), we first perform the inner function, \(g(x)\), and then use its output as the input for the outer function, \(f(x)\). When given a specific value like 2, we simply plug that number into our already composed function and calculate the result, as seen in the provided steps of the exercise.