Problem 34 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The value of \((s+t)(3)\) is 0.

Step 1: Understand the Function Addition

The expression \((s+t)(x)\) represents the sum of two functions: \(s(x)\) and \(t(x)\). Therefore, \((s+t)(x) = s(x) + t(x)\). Our goal is to find \((s+t)(3)\), which means we'll first need to find \(s(3)\) and \(t(3)\), and then sum them.

Step 2: Evaluate the Function s at 3

Substitute \(x=3\) into the function \(s(x) = 3 - x\). So, \(s(3) = 3 - 3 = 0\).

Step 3: Evaluate the Function t at 3

Substitute \(x=3\) into the function \(t(x) = x^2 - x - 6\). Calculate \(t(3) = 3^2 - 3 - 6 = 9 - 3 - 6 = 0\).

Step 4: Add the Function Values

Now that we have both \(s(3) = 0\) and \(t(3) = 0\), we sum them: \((s+t)(3) = s(3) + t(3) = 0 + 0 = 0\).

Key Concepts

Evaluating Functions

Understanding how to evaluate functions is an essential part of studying algebra. Evaluating a function means finding the output value for a given input value.
Functions take an input, which we can call "x," and process it through a formula to produce an output, often called "f(x)." To evaluate a function at a specific value, we substitute the given number in place of the variable within the function's formula.
Let's look at a couple of important points when it comes to evaluating functions:

• Identify the function's formula and the variable you are working with.
• Substitute the specific value for the variable in the function's formula.
• Simplify the expression if needed to find the output value.

This process is crucial in various applications, from physics to economics, where you need to forecast outcomes based on specific conditions.

Polynomial Functions

Polynomial functions are a type of function that includes terms with variables raised to whole number powers and different coefficients.
They can vary in complexity from simple linear polynomials, like "defining" linear polynomials with constants, to more complex quadratic, cubic, or higher-degree polynomials.
For example, the polynomial function given in our exercise is:

• Quadratic: Consists of a square term, which is why it is also called a second-degree polynomial.
• General form: A quadratic polynomial is represented as \( ax^2 + bx + c \), where a, b, and c are constants.
• Behavior: Polynomial functions have distinct characteristics depending on their degree, such as U-shaped graphs for quadratic functions.

Recognizing and manipulating these forms is crucial as they appear frequently in algebraic problems and can model real-world situations.

Function Evaluation at Specific Values

To evaluate a function like \((s+t)(x)\) at a specific value, you must perform function addition and then substitute the given value into the resultant function.
In the exercise example, you perform the following steps:

• Addition of Functions: First combine the functions \(s(x)\) and \(t(x)\) to create a new function that represents their sum.
• Substitution: Plug in the specific value—in this case, 3—into each function individually before summing the results.
• Completion: By substituting separately into both \(s(x)\) and \(t(x)\), then adding up their results, you find the combined value at that specific point.

This technique allows us to evaluate composite functions, sharpening our ability to tackle more advanced algebra problems.