Problem 59 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$x^{3}>x^{2}$$

Step-by-Step Solution

Answer

The solution is the interval \((1, fty)\).

Step 1: Write the inequality in standard form

Start by rearranging the inequality so that one side equals zero. The inequality given is \(x^3 > x^2\). Subtract \(x^2\) from both sides to obtain \(x^3 - x^2 > 0\).

Step 2: Factor the cubic expression

Factor out the common factor \(x^2\) from \(x^3 - x^2 = x^2(x - 1)\). So the inequality becomes \(x^2(x - 1) > 0\).

Step 3: Find critical points

The inequality \(x^2(x - 1) > 0\) has critical points where the expression equals zero. Set \(x^2(x - 1) = 0\). Solving this gives \(x^2 = 0\) or \(x - 1 = 0\), leading to critical points at \(x = 0\) and \(x = 1\).

Step 4: Test intervals between critical points

Given the critical points, test the sign of the expression \(x^2(x - 1)\) in the intervals \((-fty, 0)\), \((0, 1)\), and \((1, fty)\).- For \(x < 0\), choose \(x = -1\): \((-1)^2(-1 - 1) = 1(-2) < 0\).- For \(0 < x < 1\), choose \(x = 0.5\): \((0.5)^2(0.5 - 1) = 0.25(-0.5) < 0\).- For \(x > 1\), choose \(x = 2\): \((2)^2(2 - 1) = 4(1) > 0\).

Step 5: Identify the solution interval

From the testing, the expression \(x^2(x - 1)\) is positive on the interval \((1, fty)\), therefore the solution to the inequality \(x^3 > x^2\) is \((1, fty)\).

Key Concepts

Factoring Polynomials

In solving polynomial inequalities like the one presented here, factoring polynomials plays a crucial role. When faced with an expression like \(x^3 - x^2 > 0\), the first step is to rewrite it in such a way that involves zero on one side of the inequality. This allows us to apply factoring techniques.

To factor the expression \(x^3 - x^2\), we need to identify common factors. In this example, both terms share a common factor of \(x^2\). By factoring \(x^2\) out, we rewrite the polynomial as \(x^2(x - 1)\). This step simplifies the inequality and sets the stage for finding the critical points where the expression could potentially change its sign.

Factoring is vital as it breaks down complex polynomial expressions into simpler, more manageable parts. It's akin to breaking a large problem into smaller pieces, making it easier to solve.

Critical Points

Critical points are the specific values of \(x\) that make the factored polynomial equate to zero. For the inequality \(x^2(x - 1) > 0\), critical points occur when \(x^2(x - 1) = 0\).

To find these points, set each factor to zero. Start with \(x^2 = 0\), which gives \(x = 0\). Then, solve \(x - 1 = 0\) to get \(x = 1\). Thus, our critical points are at \(x = 0\) and \(x = 1\).

Identifying critical points is essential because these are potential spots where the polynomial shifts from positive to negative or vice versa. They provide boundaries for testing the sign of the expression in different intervals on the number line.

Interval Notation

Interval notation is a way of expressing the solution set of an inequality by denoting the range of values that make the inequality true. Once critical points are identified, we assess how the expression behaves in the different sections of the number line these points create.

Consider the intervals:

• \((-\infty, 0)\)
• \((0, 1)\)
• \((1, \infty)\)

The goal is to determine for which intervals the factored expression is positive, revealing where the original inequality holds. In this example, after evaluating the expression in each segment, we find that \(x^2(x - 1) > 0\) is true in the interval \((1, \infty)\).

Choosing the correct interval ensures you identify all the solutions to the inequality, offering a concise and clear way to present your findings.

Testing Intervals

Testing intervals involves checking the sign of the polynomial within the ranges demarcated by the critical points. This strategy confirms which intervals satisfy the inequality. After factoring and identifying critical points, you're left with distinct intervals on the number line.

In practice, choose a test point (any number within the interval), substitute it back into the expression, and determine the result's sign.

• For \((-\infty, 0)\), test with \(x = -1\) and find it results in a negative value.
• For \((0, 1)\), use \(x = 0.5\) and also obtain a negative result.
• For \((1, \infty)\), test with \(x = 2\), which yields a positive value.

Therefore, the portion of the number line where the expression is positive (satisfying the original inequality) is \((1, \infty)\). Testing intervals is a critical step in narrowing down the range of solutions systematically.