Problem 28 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Find \(f(x)\) for the indicated values of \(x\), if possible. (b) Find the domain of \(f\). $$ f(x)=x^{2}-x+1 \text { for } x=1,-2 $$

Step-by-Step Solution

Answer

(a) \( f(1) = 1 \) and \( f(-2) = 7 \). (b) The domain is all real numbers, \( \mathbb{R} \).

Step 1: Evaluate f(x) for x=1

To find the value of the function at \( x = 1 \), substitute 1 into the function: \[ f(1) = 1^2 - 1 + 1 \]. Simplify the expression: \( f(1) = 1 - 1 + 1 = 1 \). Thus, \( f(1) = 1 \).

Step 2: Evaluate f(x) for x=-2

To find the value of the function at \( x = -2 \), substitute -2 into the function: \[ f(-2) = (-2)^2 - (-2) + 1 \]. Simplify the expression: \( (-2)^2 = 4 \), so \( f(-2) = 4 + 2 + 1 = 7 \). Thus, \( f(-2) = 7 \).

Step 3: Determine the domain of f

The function \( f(x) = x^2 - x + 1 \) is a polynomial function. Polynomial functions are defined for all real numbers. Therefore, the domain of \( f \) is all real numbers, \( \mathbb{R} \).

Key Concepts

Function Evaluation

Evaluating a function essentially means finding out what the output of the function is for specific input values. It is like inputting a number into a machine and getting an output. For polynomial functions like the one we're dealing with here, you substitute a given value for the variable in the function and simplify.

Let's see the process clearly:

• Substitute the value of the variable into the function.
• Simplify the equation to find the result.
• In the case of polynomial functions like \( f(x) = x^2 - x + 1 \), this involves basic arithmetic operations such as addition, subtraction, and exponentiation.

In our example, when evaluating \( f(x) \) at \( x=1 \), which means substituting 1 in place of \( x \), you do the following: \( f(1) = 1^2 - 1 + 1 = 1 \). This shows that when \( x = 1 \), the output of the function is 1.

Similarly, for \( x = -2 \), substitute -2 into the function: \( f(-2) = (-2)^2 - (-2) + 1 = 7 \). So, for this input, the function outputs 7. This step-by-step substitution and simplification make it clear how function evaluation works for specific values.

Domain of a Function

The domain of a function is an important concept in mathematics. It tells us all the possible input values (often \( x \)) that we can use in the function that will give a real output. For some functions, the domain can be limited due to restrictions like division by zero or taking square roots of negative numbers.

However, when it comes to polynomial functions, like \( f(x) = x^2 - x + 1 \), the domain is quite straightforward. Polynomial functions consist of terms with variables raised to whole number powers. These functions are continuous and defined for every real number.

As a result, you don't have to worry about any restrictions or undefined points. Therefore, the domain of any polynomial function, including the given one, is all real numbers, \( \mathbb{R} \). This means no matter what real number you choose to substitute into the polynomial, it will always produce a valid output.

Real Numbers

Real numbers are the foundation of almost all the math you learn in school. They include all the numbers you could find on the number line, which means every number that isn't imaginary. More specifically, real numbers encompass:

• Natural numbers: 1, 2, 3, ...
• Whole numbers: 0, 1, 2, 3, ...
• Integers: ..., -3, -2, -1, 0, 1, 2, 3, ...
• Rational numbers: Any number that can be expressed as a fraction of two integers, like \( \frac{1}{2} \) or \( -3 \).
• Irrational numbers: Numbers that cannot be expressed as simple fractions, like \( \pi \) or \( \sqrt{2} \).

When we talk about the domain of functions or evaluate polynomials, we are typically working within the set of real numbers. This is because all calculations that use addition, subtraction, multiplication, and division on real numbers will yield more real numbers, keeping us within a very predictable and wide range of values for solving equations.