Problem 60 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\left(x^{2}-x\right)\left(x^{2}-5 x+6\right)<0$$

Step-by-Step Solution

Answer

The solution is: \((-\infty, 0) \cup (0, 1) \cup (2, 3)\).

Step 1: Factor the terms

First, we factor each quadratic expression. The expression \(x^2 - x\) can be factored as \(x(x - 1)\). The quadratic \(x^2 - 5x + 6\) can be factored as \((x - 2)(x - 3)\). This makes the inequality:\[(x)(x - 1)(x - 2)(x - 3) < 0\]

Step 2: Identify critical points

We identify the points where any factor might change sign, these are called critical points. Here, the critical points are solutions to each expression equal to zero: \(x = 0, 1, 2,\) and \(3\).

Step 3: Determine sign changes between critical points

These critical points divide the number line into intervals: \((-fty, 0)\), \((0, 1)\), \((1, 2)\), \((2, 3)\), and \((3, fty)\). We now check the sign of the product \((x)(x - 1)(x - 2)(x - 3)\) in each interval by choosing a test point from each interval. For instance, for the interval \((-fty, 0)\), test \(x = -1\), and so forth.

Step 4: Calculate the signs of each interval

Calculate the signs:- \((-fty, 0)\) : Choose \(x = -1\), signs are negative overall (\< 0).- \((0, 1)\) : Choose \(x = 0.5\), signs are negative overall (\< 0).- \((1, 2)\) : Choose \(x = 1.5\), signs are positive (\> 0).- \((2, 3)\) : Choose \(x = 2.5\), signs are negative overall (\< 0).- \((3, fty)\) : Choose \(x = 4\), signs are positive (\> 0).

Step 5: Combine intervals with negative products

Based on the calculated signs, the inequality \((x)(x - 1)(x - 2)(x - 3) < 0\) is satisfied on intervals where the product is negative: \((-\infty, 0)\), \((0, 1)\), and \((2, 3)\).

Step 6: Write the solution in interval notation

The solution to the inequality \(x^2 - x)(x^2 - 5x + 6) < 0\) is expressed in interval notation as: \((-\infty, 0) \cup (0, 1) \cup (2, 3)\).

Key Concepts

Factoring Quadratics

Factoring quadratics is a key step in solving quadratic inequalities. Just like breaking down a complex machine into simpler parts, factoring helps you see each component clearly. The process involves expressing the quadratic polynomial as a product of its factors. For example, with the expressions provided:

• The quadratic expression \(x^2 - x\) factors into \(x(x - 1)\). This means finding numbers that, when multiplied, give you the original expression and add up to the coefficient of the middle term (if present).
• Similarly, \(x^2 - 5x + 6\) factors into \((x - 2)(x - 3)\). These factors come from finding numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of \(x\)).

When you factor quadratics, solving gets much easier because you can work with simpler linear expressions. Factoring sets the stage for finding solutions by zeroing in on values that make each factor zero—our critical points.

Critical Points

Critical points in the context of inequalities are values where a polynomial changes its sign. These points occur where the expression equals zero. Once a quadratic has been factored, finding its critical points is straightforward. Simply set each factor equal to zero:

• For \(x(x - 1)(x - 2)(x - 3)\), solve each factor:
• \(x = 0\)
• \(x - 1 = 0\), so \(x = 1\)
• \(x - 2 = 0\), so \(x = 2\)
• \(x - 3 = 0\), so \(x = 3\)

These critical points \(x = 0, 1, 2,\) and \(3\) help in determining the sign of expressions before and after these values. They'll also divide the number line into intervals that can be individually analyzed.

Interval Notation

Interval notation is like a shorthand method for describing sets of numbers. It provides a convenient way of representing the intervals that satisfy an inequality. In this method:

• An open interval, such as \((a, b)\), includes all numbers greater than \(a\) and less than \(b\), but not \(a\) or \(b\) themselves.
• A closed interval, written as \([a, b]\), includes \(a\), \(b\), and all the numbers in between.
• If an endpoint extends to infinity, like \((-fty, a)\) or \((b, fty)\), it means the interval extends endlessly in the specified direction.

Interval notation is particularly handy in expressing the solution to inequalities. For the inequality \((x)(x - 1)(x - 2)(x - 3) < 0\), the solution is expressed as \((-fty, 0) \cup (0, 1) \cup (2, 3)\). Here, "\(\cup\)" signifies the union of intervals, meaning any number in these intervals solves the inequality.

Sign Analysis

Sign analysis is a critical part of solving inequalities. The goal is to determine whether the product expression \((x)(x - 1)(x - 2)(x - 3)\) is positive or negative in each identified interval. You can achieve this by choosing test points within each interval and plugging them into the factored expression:

• For \((-fty, 0)\), pick a point like \(x = -1\). Calculate the sign of the expression, and you'll find it negative.
• Continue with other intervals: \((0, 1)\), choose \(x = 0.5\); \((1, 2)\), choose \(x = 1.5\); \((2, 3)\), choose \(x = 2.5\); and \((3, fty)\), choose \(x = 4\).

Each test reveals whether the full expression \((x)(x - 1)(x - 2)(x - 3)\) is greater or less than zero in those intervals. The solution to the inequality involves combining intervals where the product is negative: \((-fty, 0)\), \((0, 1)\), and \((2, 3)\). By confirming which segments have the desired inequality property, you map out the solution visually and logically.