Problem 28 - Precalculus

Question

Solve the equation or inequality. $$\sqrt[3]{x} \leq x$$

Step-by-Step Solution

Answer

Solution is \([-1, 0) \cup [1, \infty)\).

Step 1: Define the Equation

We start with the inequality \(\sqrt[3]{x} \leq x\). This means the cube root of \(x\) should be less than or equal to \(x\) itself.

Step 2: Analyze Sign of x

The inequality might behave differently on different intervals of \(x\). Consider cases: when \(x > 0\), \(x = 0\), and \(x < 0\).

Step 3: Solve for Positive x

For \(x > 0\): Because both sides of the inequality are positive, raise both sides to the power of 3: \[x \leq x^3\]This simplifies to \(x^3 - x \geq 0\).

Step 4: Factor the Inequality

Factor \(x^3 - x\):\[x(x^2 - 1) \geq 0\]which is further factored as \(x(x - 1)(x + 1) \geq 0\).

Step 5: Test Intervals

Use a sign chart or test points in the intervals created by the roots \(-1, 0, 1\) to determine where the inequality holds:- Interval \((-\infty, -1)\) : Negative- Interval \((-1, 0)\) : Positive- Interval \((0, 1)\) : Negative- Interval \((1, \, \infty)\) : Positive

Step 6: Solution for x > 0 and x < 0

Combine positive interval solutions from test values for entire \(x\) range:- From \((0, 1)\), \(x = 0\) does not work (but \(x = 0\) fits original equation, keep it)- From \((1, \, \infty)\), works.Overall, solution includes \([-1, 0) \cup [1, \infty)\).

Key Concepts

Cube Roots

The concept of cube roots is essential when solving equations or inequalities involving cubic expressions. A cube root of a number \(x\) is a number \(y\) such that \(y^3 = x\). In our inequality \(\sqrt[3]{x} \leq x\), we compare the cube root of \(x\) to \(x\) itself.

• Understanding cube roots helps simplify and solve complex inequalities or equations.
• Cube roots, unlike square roots, can be negative because a negative number raised to the third power will result in a negative number.
• This allows us to work with negative values of \(x\) in our inequality.

When dealing with cube roots, we often need to consider different cases for \(x\), such as when \(x\) is positive, negative, or zero, since each can affect the outcome of the inequality. Understanding the cube root's behavior helps us to understand which power or function might be higher.

Interval Notation

Interval notation is a concise method to describe a set of numbers on a number line. It is particularly useful when specifying solutions to inequalities, like \(\sqrt[3]{x} \leq x\).

• Closed intervals \([a, b]\) include the endpoints \(a\) and \(b\), meaning both values are part of the solution set.
• Open intervals \((a, b)\) do not include the endpoints, indicating the values in between \(a\) and \(b\) are part of the solution set but not \(a\) or \(b\) themselves.
• When using interval notation, combinations such as \([-1, 0) \cup [1, \infty)\) are allowed, which combine different intervals to make the complete solution set.

The notation \([-1, 0) \cup [1, \infty)\) means all numbers from \(-1\) to 0, excluding 0, and all numbers from 1 upwards. This provides clarity in presenting results of inequalities without listing infinite values.

Sign Chart

A sign chart is a valuable tool when determining where a function is positive or negative over certain intervals. In the inequality \(x(x-1)(x+1) \geq 0\), a sign chart helps identify the intervals where the product is non-negative.

• Identify the roots of the polynomial, here \(-1\), \(0\), and \(1\), which divide the number line into distinct intervals.
• For each interval, select a test point, substitute it into the inequality, and determine the sign of the product.
• The sign indicates whether the product is positive or negative in that interval, guiding us to test intervals for solutions.

By analyzing intervals \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, \infty)\), we can determine where the inequality holds. For instance, in the interval \((1, \infty)\), the product is positive, indicating solutions exist there.

Factoring Polynomials

Factoring polynomials is crucial when solving inequalities or equations involving expressions like \(x^3 - x\). The process involves breaking down a complex polynomial into simpler factors that are easier to manage and solve.

• Start by factoring out common terms: for \(x^3 - x\), factor \(x\), resulting in \(x(x^2 - 1)\).
• Next, factor any polynomials further: \(x^2 - 1\) can be factored with the difference of squares to \((x-1)(x+1)\).
• Use these factors to form the inequality \(x(x-1)(x+1) \geq 0\).

Factoring reduces the problem to checking where the factors are positive or zero, enabling us to apply the sign chart method effectively. This decomposition of the polynomial makes solving inequalities straightforward.