Problem 54 - Precalculus Mathematics for Calculus

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$x^{2}\left(x^{2}-1\right) \geq 0$$

Step-by-Step Solution

Answer

The solution is \([-1, 0] \cup [1, \infty)\).

Step 1: Identify the Expression

The inequality is given as \(x^2(x^2 - 1) \geq 0\). This means we want to find the values of \(x\) for which this expression is greater than or equal to zero.

Step 2: Factor the Expression

Factor the expression \(x^2(x^2 - 1)\). It can be rewritten as \(x^2(x-1)(x+1)\), because \(x^2 - 1\) is a difference of squares and factors into \((x-1)(x+1)\).

Step 3: Determine Critical Points

Identify the critical points where the inequality might change sign. These are the values that make each factor zero: \(x^2 = 0\) (giving \(x = 0\)), \(x-1 = 0\) (giving \(x = 1\)), and \(x+1 = 0\) (giving \(x = -1\)).

Step 4: Test Intervals Around Critical Points

Divide the number line based on the critical points: \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, \infty)\). Test a sample point from each interval in the expression \(x^2(x-1)(x+1)\) to determine the sign.

Step 5: Analyze the Sign in Each Interval

- In \((-\infty, -1)\), choose \(x = -2\). The expression \((-2)^2((-2)-1)((-2)+1) = 4(-3)(-1) > 0\).- In \((-1, 0)\), choose \(x = -0.5\). The expression \((-0.5)^2((-0.5)-1)((-0.5)+1) = 0.25(-1.5)(0.5) < 0\).- In \((0, 1)\), choose \(x = 0.5\). The expression \((0.5)^2((0.5)-1)((0.5)+1) = 0.25(-0.5)(1) < 0\).- In \((1, \infty)\), choose \(x = 2\). The expression \((2)^2((2)-1)((2)+1) = 4(1)(3) > 0\).

Step 6: Include the Critical Points

At the critical points \(-1\), \(0\), and \(1\): - \(x^2(x-1)(x+1) = 0\) which satisfies the \(\geq 0\) condition.

Step 7: Write the Solution in Interval Notation

Include the intervals where the expression is non-negative and the critical points that satisfy the inequality. The solution set is \([-1, 0] \cup [1, \infty)\).

Step 8: Graph the Solution Set

On a number line, draw closed circles at points \(-1\) and \(0\), and draw a closed circle at \(1\) with a line extending to \(\infty\). The segments \([-1, 0]\) and \([1, \infty)\) should be highlighted, representing the solution.

Key Concepts

Interval Notation

When solving inequalities like the one in this exercise, expressing the solution with interval notation is incredibly useful. Interval notation provides a concise way to represent a range of values that satisfy an inequality. For instance, the solution

• \([-1, 0]\) indicates all real numbers between -1 and 0, inclusive.
• \([1, \infty)\) indicates all real numbers from 1 to infinity, inclusive at 1 and going on forever.

The brackets \([]\) mean that the number is included in the solution, while parentheses \(())\) mean it is not. By combining intervals, such as using the union symbol \(\cup\), we easily showcase the different parts of the solution set that satisfy the inequality.

Critical Points

Critical points are crucial in understanding where the inequality changes direction, meaning where the given expression equals zero. This guides which intervals on the number line need to be tested further. To identify critical points, set each part of the factored equation to zero:

• For \(x^2 = 0\), we get \(x = 0\).
• For \(x-1 = 0\), we find \(x = 1\).
• For \(x+1 = 0\), we derive \(x = -1\).

These values split the number line into sections we can test, giving us points where the expression changes sign between positive, negative, and zero. Thus, critical points serve as the dividing markers that help solve the inequality.

Difference of Squares

The difference of squares is a valuable factoring technique used in this problem to simplify the expression. This method is useful when dealing expressions of the form \(a^2 - b^2\), which can be factored as \((a-b)(a+b)\). In the context of the exercise, the expression \(x^2-1\) is a difference of squares:

• It can be rewritten as \((x-1)(x+1)\).

This factoring step simplifies the inequality, revealing the critical points more clearly and allowing us to solve or graph the equation efficiently. It is an essential part of solving quadratics and enhances our ability to manipulate algebraic expressions.

Solution Set

The solution set is the final answer, which includes all the values of \(x\) that satisfy the inequality. In this exercise, after determining the signs within each interval divided by the critical points, we concluded:

• The expression is non-negative in the intervals \([-1, 0]\) and \([1, \infty)\).
• At the critical points \(-1, 0,\) and \(1\), the expression equals zero, satisfying the \(\geq 0\) condition.

Putting it all together, we write the solution using interval notation: \([-1, 0] \cup [1, \infty)\). The graph of this solution on the number line highlights the segments that are part of this set, providing a visual representation of all possible solutions to the inequality. It neatly shows where on the number line the inequality holds true.