Problem 34 - Precalculus

Question

Solve the equation or inequality. $$x^{-\frac{1}{3}}(x-3)^{-\frac{2}{3}}-x^{-\frac{4}{3}}(x-3)^{-\frac{5}{3}}\left(x^{2}-3 x+2\right) \geq 0$$

Step-by-Step Solution

Answer

The inequality holds for \(0 < x < 1\), \(1 < x < 2\), and \(2 < x < 3\).

Step 1: Rewrite the Equation

Start by factoring out the common terms in each part of the inequality. Observe that both terms have common factors of \(x^{-rac{1}{3}}(x-3)^{-rac{2}{3}}\). Factor out these common terms:\[x^{-rac{1}{3}}(x-3)^{-rac{2}{3}}\left(1 - x^{-1}(x-3)^{-1}(x^2 - 3x + 2)\right) \geq 0\]Simplify the factored term.

Step 2: Simplify the Expression

Simplify \(x^{-1}(x-3)^{-1}(x^2 - 3x + 2)\) by dividing the polynomial \(x^2 - 3x + 2\) by \((x)(x-3)\):\[x^{-1}(x-3)^{-1}\left((x-1)(x-2)\right)\]Thus, the inequality becomes:\[x^{-rac{1}{3}}(x-3)^{-rac{2}{3}}(1 - x^{-1}(x-3)^{-1}(x-1)(x-2)) \geq 0\]

Step 3: Set the Inner Expression to Zero

Focus on the inner expression \(1 - x^{-1}(x-3)^{-1}(x-1)(x-2)\). Set it equal to zero and solve for \(x\):\[1 = \frac{(x-1)(x-2)}{x(x-3)}\]Cross-multiply to obtain:\[x(x-3) = (x-1)(x-2)\]Solve by expanding and rearranging:\[x^2 - 3x = x^2 - 3x + 2\]0 = 2. There is no valid solution for this inner expression being zero.

Step 4: Consider the Sign of the Expression

Since \(1 - x^{-1}(x-3)^{-1}(x-1)(x-2)\) never equals zero, examine its sign based on critical points: \(x = 0\), \(x = 3\), \(x = 1\), and \(x = 2\). These points are where factors change sign because they are critical points of \(x(x-3)\) and \((x-1)(x-2)\).

Step 5: Test Intervals Between Critical Points

Test intervals determined by critical points: \((-\infty, 0)\), \((0, 1)\), \((1, 2)\), \((2, 3)\), and \((3, \infty)\).1. For \(x \in (0, 1)\), \(x^{-1}\), \((x-3)^{-1}\), and \((1-2)\) have different signs implying overall expression positive.2. For \(x \in (1, 2)\), test to find inequality holds true.3. Repeat for other intervals. Expression is positive for \(x \in (2, 3)\).4. Note the domains due to restrictions in original terms where denominator cannot zero.

Step 6: Write the Solution

Based on test intervals, identify where the inequality holds: \((0, 1) \cup (1, 2) \cup (2, 3)\).

Key Concepts

Polynomial Inequalities

Polynomial inequalities are mathematical expressions that involve a polynomial on one or both sides of an inequality sign, such as \(\geq, \leq, >, <\). In the given exercise, the inequality is:

• \(x^{\frac{-1}{3}}(x-3)^{\frac{-2}{3}} - x^{\frac{-4}{3}}(x-3)^{\frac{-5}{3}}(x^2 - 3x + 2) \geq 0\)

At their core, these inequalities require us to determine the ranges of \(x\) for which the inequality is satisfied. They are vital in understanding behaviors of functions, especially where they don't behave linearly.
When tackling polynomial inequalities, the main goal is to

• Identify when the polynomial is greater than (or less than) zero.
• Consider regions where the polynomial changes sign and recognize any restrictions.

This process is essential to explore the positive or negative nature of polynomials, especially around critical points, which we will cover later.

Factoring Techniques

Factoring techniques are crucial for simplifying and solving polynomial inequalities. To factor an expression means to break it down into products of simpler expressions that, when multiplied together, yield the original polynomial. Factoring reveals zeros and allows easier manipulation of equations. In the original problem, the equation is factored by:

• Identifying common components such as \(x^{\frac{-1}{3}}(x-3)^{\frac{-2}{3}}\).
• Expressing more complicated terms as simpler fractions or products.

By factoring, you turn what seems like a complex equation into a more manageable form, structured around its roots. Another key step is removing these common factors, simplifying not just the equation structure, but also focusing on core expressions that impact the inequality. Factoring plays a pivotal role as the groundwork for further strategies like Interval Testing.

Interval Testing

Interval testing is a technique used in many inequality problems to determine where an inequality holds true. Once the polynomial is factored, like in the current case, interval testing comes into play to evaluate which segments satisfy the inequality.

• Examine between critical values, often where the inequality sign might change.
• Test simple points within each interval to determine positivity or negativity.

For example, consider intervals created by the critical points \(0, 1, 2,\) and \(3\).
By picking test points, such as 0.5 in \((0, 1)\), you determine if the entire segment is in the solution set. These test points reveal the overall sign of the combined expression, allowing you to map where the inequality is satisfied across the number line.
Therefore, interval testing provides a straightforward method to solve and validate the range of solutions.

Critical Points Analysis

Critical Points Analysis involves identifying points at which the polynomial or component expressions are undefined or zero. It's necessary for parsing where factors influence the sign of the whole inequality. In the given problem, critical points include \(x = 0, 1, 2,\) and \(3\), as they are places where the equation's terms can equal zero or become undefined, especially considering the denominators.

• These points highlight where the function can shift from positive to negative or vice versa.
• Tapping into these points helps unveil the behavior of the function around essential junctions.

Understanding critical points allows for determining where specific values cause disconnection or zero in your inequality's domain. It leads directly into the testing for positivity status to determine the satisfied intervals of your inequality.
Thus, critical points form the backbone to ascertain which sections between them fulfill or contradict the given inequality constraints.