Problem 31 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

\((f \cdot g)(0) = -5\).

Step 1: Understand Function Multiplication

The notation \((f \cdot g)(x)\) means to multiply the functions \(f(x)\) and \(g(x)\). This means we will find \(f(x) \cdot g(x)\) and evaluate it at \(x = 0\).

Step 2: Write Down the Functions

We are given the functions \(f(x) = 2x - 5\) and \(g(x) = x + 1\). We need to find the product \(f(x) \cdot g(x)\).

Step 3: Multiply the Functions

Multiply \(f(x)\) by \(g(x)\) as follows:\[f(x) \cdot g(x) = (2x - 5)(x + 1)\].

Step 4: Expand the Expression

Use the distributive property to expand the expression:\[(2x - 5)(x + 1) = 2x \cdot x + 2x \cdot 1 - 5 \cdot x - 5 \cdot 1\]Simplifying each term, we get:\[2x^2 + 2x - 5x - 5\]Combine like terms:\[2x^2 - 3x - 5\].

Step 5: Evaluate at \(x=0\)

Substitute \(x = 0\) into the expression \(2x^2 - 3x - 5\):\[2(0)^2 - 3(0) - 5 = 0 - 0 - 5 = -5\].

Step 6: Write the Final Answer

Thus, \((f \cdot g)(0) = -5\).

Key Concepts

Function Multiplication

When we talk about function multiplication, it means you are multiplying two or more functions together. In this exercise, we're looking at two functions: \(f(x) = 2x - 5\) and \(g(x) = x + 1\). To multiply these functions, you take the entire expression of \(f(x)\) and multiply it by the entire expression of \(g(x)\). Think of it as expanding a multiplication between two binomials. For \((f \cdot g)(x)\), you will perform \([2x - 5]\) times \([x + 1]\), which invites us to use the distributive property to expand and simplify this expression. By understanding function multiplication, you set the stage for diving deeper into algebraic manipulation and prepare for integrating functions in calculus.

Distributive Property

The distributive property is crucial when multiplying functions, especially those in polynomial form. This property allows us to expand expressions such as \((2x - 5)(x + 1)\). The distributive property states that for any values of \(a\), \(b\), and \(c\), the equation \(a(b + c) = ab + ac\) holds.Therefore, when you expand \((2x - 5)(x + 1)\), you distribute each term in the first binomial through each term in the second binomial:

• First, multiply \(2x\) by \(x\) to get \(2x^2\).
• Then, multiply \(2x\) by \(1\) to get \(2x\).
• Next, multiply \(-5\) by \(x\) to get \(-5x\).
• Finally, multiply \(-5\) by \(1\) to get \(-5\).

This results in the expression: \(2x^2 + 2x - 5x - 5\). Combine like terms to further simplify to \(2x^2 - 3x - 5\). By mastering the distributive property, you can confidently tackle function multiplications and similar algebraic operations.

Evaluation of Functions

Evaluating a function means finding the value of the function for a certain value of \(x\). After finding \(f(x) \cdot g(x) = 2x^2 - 3x - 5\), the next step is evaluation at \(x = 0\). Substitute \(0\) into the expression and simplify:

• Replace \(x\) with \(0\) in \(2x^2 - 3x - 5\).
• Calculate each term: \(2(0)^2 = 0\), \(-3(0) = 0\), and \(-5\) remains \(-5\).

Add these results to find that the value of the expression at \(x = 0\) is \(-5\). Through function evaluation, you directly determine function values at specific points, a fundamental skill in both algebra and calculus. This process helps verify your solutions and understand the behavior of functions graphically and numerically.