Problem 31 - Elementary and Intermediate Algebra

Question

Let \(f(x)=2 x-5\) and \(g(x)=x+1 .\) Find each of the following function values. $$ (f \cdot g)(0) $$

Step-by-Step Solution

Answer

The value of \((f \cdot g)(0)\) is -5.

Step 1: Understand the operation needed

The exercise asks for the value of \((f \cdot g)(0)\). This indicates that you must first find the product of the functions \(f(x)\) and \(g(x)\), and then evaluate this product at \(x=0\).

Step 2: Find the product of the functions

To find the product of the functions, multiply the expressions for \(f(x)\) and \(g(x)\):\[f(x) = 2x - 5\]\[g(x) = x + 1\]Thus, \[(f \cdot g)(x) = (2x - 5)(x + 1)\]

Step 3: Expand the product expression

Use the distributive property to expand the expression \((2x - 5)(x + 1)\):\[(2x - 5)(x + 1) = 2x(x) + 2x(1) - 5(x) - 5(1)\]Simplifying, we have:\[= 2x^2 + 2x - 5x - 5 = 2x^2 - 3x - 5\]

Step 4: Evaluate the product at x=0

We now substitute \(x=0\) into the expanded expression:\[(2x^2 - 3x - 5)_{x=0} = 2(0)^2 - 3(0) - 5\]Simplifying, we get:\[= 0 - 0 - 5 = -5\]

Key Concepts

Algebraic Functions

Algebraic functions are mathematical expressions that involve variables and constants combined using algebraic operations like addition, subtraction, multiplication, and division of polynomials. They play a crucial role in understanding relationships between variables. Some key points about algebraic functions include:

• They can be simple, such as linear functions like \(f(x) = 2x - 5\), or more complex involving multiple terms and higher powers.
• The different forms of algebraic functions allow us to model real-world situations, ranging from calculating interests to analyzing motion.
• Combining these functions via operations (like addition or multiplication) can help in solving more comprehensive problems or analyses.

Understanding the nature of these functions is foundational in algebra, as it lays the groundwork for more advanced topics like calculus. In this exercise, you see functions \(f(x) = 2x - 5\) and \(g(x) = x + 1\), both are linear expressions representing straight lines when plotted on a graph.

Function Evaluation

Function evaluation refers to the process of determining the output of a function for a specific input value. In simpler terms, it's about plugging a number into the function to find out what the function's result is at that point. Here's what's essential to know:

• To evaluate a function, simply substitute the given value for the variable \(x\) in the function expression, and then simplify.
• This process helps to find the specific result for given inputs, which is vital in verifying and predicting outcomes in mathematical models.
• Encountering different points of the function provides insights into the behavior or trends of the function's graph.

In the given problem, the expression \((f \cdot g)(0)\) means to find the product of the functions at \(x = 0\). After expanding the product expression to \(2x^2 - 3x - 5\), substituting \(x = 0\) gives us the precise result of \(-5\). This shows how inputs influence the function outcome.

Distributive Property

The distributive property is a powerful algebraic property used to multiply a sum by distributing the multiplication over individual terms inside the parentheses. It consolidates expressions and is essential for expanding products of algebraic expressions.

• Mathematically, it is expressed as \(a(b + c) = ab + ac\).
• Using the distributive property simplifies expressions by eliminating parentheses, which helps in further calculations.
• It's also used to factor back simplified expressions into their original multiplied form.

In our exercise, the distributive property was applied to expand \((2x - 5)(x + 1)\). By distributing each term from the first polynomial to each term in the second, we can reach the expanded form \(2x^2 - 3x - 5\). This simplification is crucial for evaluating the function at specific values, such as \(x = 0\) in this case.