Problem 34 - Intermediate Algebra

Question

Let \(s(x)=3-x\) and \(t(x)=x^{2}-x-6 .\) Find each function value. See Example 2 . $$ (s+t)(3) $$

Step-by-Step Solution

Answer

(s+t)(3) = 0.

Step 1: Understand the Problem

We are given two functions: \(s(x) = 3 - x\) and \(t(x) = x^2 - x - 6\). We need to find the value of \((s+t)(3)\). This means we need to find \(s(3) + t(3)\).

Step 2: Calculate s(3)

To find \(s(3)\), substitute \(x = 3\) into the function \(s(x) = 3 - x\):\[s(3) = 3 - 3 = 0.\]

Step 3: Calculate t(3)

To find \(t(3)\), substitute \(x = 3\) into the function \(t(x) = x^2 - x - 6\):\[t(3) = 3^2 - 3 - 6 = 9 - 3 - 6 = 0.\]

Step 4: Find (s+t)(3)

Combine the results from Steps 2 and 3 to find \((s+t)(3)\):\[(s+t)(3) = s(3) + t(3) = 0 + 0 = 0.\]

Key Concepts

Function Evaluation

Function evaluation is the process of finding the value of a function for a particular input. When you're given a function, such as \(f(x)\), and you want to find \(f(a)\), you simply replace the variable \(x\) with the number \(a\) in the expression.
This concept allows us to examine how functions behave with specific inputs and is essential for solving problems in algebra.

• To evaluate a function, identify the expression of the function and the value you want to substitute.
• Replace every occurrence of the variable with the given number.
• Perform the arithmetic operations to obtain the result.

Using the example from the exercise, to evaluate \(s(3)\), we substitute \(x=3\) into the function \(s(x) = 3-x\). Doing the arithmetic, we find \(s(3) = 3-3 = 0\).
Similarly, for \(t(x)\), replacing \(x=3\) gives \(t(3) = 3^2 - 3 - 6\), resulting in \(t(3) = 0\). This step ensures we understand how to manipulate and calculate with specific numbers in functions.

Polynomial Functions

A polynomial function is a type of mathematical expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents.
Polynomial functions are foundational in algebra and provide a broad framework for solving equations.Each polynomial function can be broken down into several terms, each consisting of a coefficient and a variable raised to a power. Let's look at the function \(t(x) = x^2 - x - 6\).

• It contains three terms: \(x^2\), \(-x\), and \(-6\).
• The highest exponent of the variable \(x\) in this function is 2, making it a quadratic polynomial.
• The degree of a polynomial is determined by the highest power of the variable, which helps classify the function.

Understanding polynomials is crucial as they frequently appear in equations and are used as a model for various real-world phenomena.

Function Addition

Function addition involves summing two or more functions together. This is done by adding their respective expressions.
In doing so, you create a new function that combines the influences of the original functions.For instance, consider the functions \(s(x) = 3-x\) and \(t(x) = x^2 - x - 6\). To add these functions, you would simply add their expressions:\[(s+t)(x) = (3-x) + (x^2-x-6)\]

• Combine like terms: \(x^2 -2x - 3\).
• This is your new function, \((s+t)(x)\).
• Function addition is straightforward, as it involves common algebraic rules and combines the effects of different functions.

In the original exercise, we specifically evaluated this addition at \(x=3\) by finding the sum of \(s(3)\) and \(t(3)\), leading to the solution of \((s+t)(3) = 0\). Function addition helps in assessing the combined impact or net effect of functions.