Problem 7 - College Algebra with Modeling and Visualization

Question

Exercises \(7-10:\) Use \(f(x)\) and \(g(x)\) to evaluate each expression symbolically. \(f(x)=2 x-3, g(x)=1-x^{2}\) (a) \((f+g)(3)\) (b) \((f-g)(-1)\) (c) \((f g)(0)\) (d) \((f / g)(2)\)

Step-by-Step Solution

Answer

(a) -5, (b) -5, (c) -3, (d) -1/3.

Step 1: Understanding the Problem

We are given two functions, \( f(x) = 2x - 3 \) and \( g(x) = 1 - x^2 \). The task is to evaluate the expressions for (a) \((f+g)(3)\), (b) \((f-g)(-1)\), (c) \((fg)(0)\), and (d) \((f/g)(2)\). This involves substituting specific values into the sum, difference, product, and quotient of the functions.

Step 2: Evaluate (f+g)(3)

First, we need to find \((f+g)(x)\) which is the additive combination: \( (f+g)(x) = f(x) + g(x) = (2x-3) + (1-x^2) \). Simplifying this, we get: \( f(x) + g(x) = -x^2 + 2x - 2 \). Now, substitute \(x = 3\): \((f+g)(3) = -(3)^2 + 2(3) - 2 = -9 + 6 - 2 = -5 \).

Step 3: Evaluate (f-g)(-1)

Next, find \((f-g)(x)\), which is the subtractive combination: \( (f-g)(x) = f(x) - g(x) = (2x-3) - (1-x^2) \). Simplifying: \( 2x - 3 - 1 + x^2 = x^2 + 2x - 4 \). Substitute \(x = -1\): \((f-g)(-1) = (-1)^2 + 2(-1) - 4 = 1 - 2 - 4 = -5 \).

Step 4: Evaluate (fg)(0)

For \((fg)(x)\), multiply the functions: \( (fg)(x) = f(x) \cdot g(x) = (2x-3)(1-x^2) \). Evaluate directly by substituting \(x = 0\): \((fg)(0) = (2(0)-3)(1-(0)^2) = -3 \cdot 1 = -3 \).

Step 5: Evaluate (f/g)(2)

For the division \((f/g)(x)\), \((f/g)(x) = \frac{f(x)}{g(x)} = \frac{2x-3}{1-x^2} \). First, substitute \(x = 2\) into \(f(x)\) and \(g(x)\): \(f(2) = 2(2) - 3 = 4 - 3 = 1\) and \(g(2) = 1 - (2)^2 = 1 - 4 = -3\). Therefore, \((f/g)(2) = \frac{1}{-3} = -\frac{1}{3}\).

Key Concepts

Algebraic Functions

Algebraic functions are expressions composed of arithmetic operations such as addition, subtraction, multiplication, and division, involving variables and constants. A function assigns a single output for each input from its domain. These operations on functions can include combining two or more functions in various ways to form new functions.
In the given exercise, we have two specific algebraic functions:

• \( f(x) = 2x - 3 \)
• \( g(x) = 1 - x^2 \)

Understanding how to manipulate these functions through different operations is fundamental to solving problems in algebra. These operations essentially allow us to explore how the behavior of both functions can affect one another when they are combined.

Symbolic Evaluation

Symbolic evaluation involves assessing expressions by substituting specific values into the variables of a function. In simpler terms, it's like plugging numbers into a formula and seeing what you get.
Let's break down the process using our functions:

• First, substitute the chosen numbers into the expression.
• Simplify the expression step by step.

For instance, in the exercise, symbolic evaluation is used to find the results for expressions like

• \((f+g)(3)\)
• \((f-g)(-1)\)

In each case, you replace \(x\) with the given number and then perform the operations as indicated. This process helps verify the function's behavior at specific points, offering insights into the result of different operations on the algebraic functions.

Function Combination

Function combination refers to the process of creating new functions by combining existing ones through mathematical operations. Here are several ways functions can be combined:

• Addition: Combine two functions by adding their expressions together. For example, \((f+g)(x) = f(x) + g(x)\) becomes \(-x^2 + 2x - 2\) when simplified.
• Subtraction: Subtract one function from another. As seen in \((f-g)(x) = f(x) - g(x)\), leading to \(x^2 + 2x - 4\) after simplification.
• Multiplication: Multiply the expressions of two functions. For example, \((fg)(x) = f(x) \cdot g(x)\), simplifies the example to a single expression concerning both \(f(x)\) and \(g(x)\).
• Division: Divide one function by another to create a quotient. In \((f/g)(x) = \frac{f(x)}{g(x)}\), be cautious as division by zero is undefined.

Exploring how functions work together allows for a deeper mathematical understanding and abstraction within algebra, facilitating more complex problem solving and expression simplification.