Problem 54 - Precalculus : Building Concepts and Connections

Question

Find polynomials \(p(x)\) and \(q(x),\) with \(q(x)\) not a constant function, such that \(\frac{p(x)}{q(x)} \geq 0\) has the solution set \([3, \infty)\) There may be more than one correct answer.

Step-by-Step Solution

Answer

Polynomials that satisfy the given conditions are \(p(x) = (x-3)^2\) and \(q(x) = 1\). However, this is not unique and there may be more than one correct answer. The solution set of the inequality \(\frac{p(x)}{q(x)} \geq 0\) is indeed \([3, \infty)\) for these polynomials.

Step 1: Define the Polynomials

We start by defining the polynomial function to satisfy our conditions. A simple polynomial that takes the value \(0\) at \(x=3\) and is positive for \(x > 3\) is \(p(x) = (x-3)^2\). This meets the criteria for \(p(x)\), since it equals zero when \(x = 3\) and is positive for \(x > 3\). For \(q(x)\), we can take just a simple non-constant polynomial that is always positive. In this case, we might use \(q(x) = 1\). This function satisfies the requirement that \(q(x)\) not be a constant function.

Step 2: Check the Solution

We test the defined \(p(x)\) and \(q(x)\) to make sure they fit the condition. By substituting the defined \(p(x)\) and \(q(x)\) into the inequality \(\frac{p(x)}{q(x)} \geq 0\), the inequality becomes \(\frac{(x-3)^2}{1} \geq 0\) for \(x \geq 3\). However, we know that \((x-3)^2\) is always nonnegative and equals zero at \(x = 3\), and hence the inequality holds true and the solution set is \([3, \infty)\). Therefore, the defined polynomials meet our condition, and there may also be other polynomials that satisfy the condition.

Key Concepts

Polynomials

Polynomials are mathematical expressions that consist of variables and coefficients, constructed using operations of addition, subtraction, and multiplication. A polynomial takes the general form:

• Constant Polynomial: Only a constant number (e.g., 5).
• Linear Polynomial: Contains a variable raised to the first power (e.g., 3x + 1).
• Quadratic Polynomial: Involves a variable squared (e.g., 2x^2 + 5x - 3).

For the exercise, we need a polynomial, such as \(p(x) = (x-3)^2\), that is zero at a certain point and positive afterward. The expression \((x-3)^2\) results in zero when \(x=3\), and is positive for any \(x > 3\), which perfectly synchronizes with our requirements.

Solution Sets

In mathematics, the concept of solution sets arises when solving equations or inequalities. A solution set includes all the values of the variable that satisfy the condition of the given problem.
For instance:

• If an equation is given as \(x + 2 = 5\), the solution set is \(\{3\}\), since \(x = 3\) satisfies the equation.
• For inequalities, such as \(x > 1\), the solution set can be an interval, in this case, \((1, \infty)\).

For the original exercise, the solution set \([3, \infty)\) indicates the range of \(x\) values where the inequality \(\frac{p(x)}{q(x)} \geq 0\) holds true. This interval captures all \(x\) starting from 3 and extending to infinity.

Inequalities

Inequalities are used to express relationships where quantities are not equal, and are indicated using symbols such as \(>\), \(<\), \( \geq \), and \( \leq \). They can define ranges of values rather than specific ones, which is particularly useful in real-world applications.
The inequality \(\frac{p(x)}{q(x)} \geq 0\) in the exercise involves two polynomials, where \(q(x)\) isn't a constant. In solving rational inequalities like this one, it's essential to understand the behavior of the polynomials involved to identify valid solution sets.

• A rational function is in the form \(\frac{p(x)}{q(x)}\), where the function is undefined wherever \(q(x)=0\).
• The inequality is true when the rational function is non-negative, i.e., greater than or equal to zero.

In particular, our function \((x-3)^2/1\) is non-negative for \(x \geq 3\), aligning perfectly with the sought solution set \([3, \infty)\).