Problem 60 - Algebra and Trigonometry

Question

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{3}{x-1}-\frac{4}{x} \geq 1 $$

Step-by-Step Solution

Answer

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

Step 1: Establish Critical Points

Critical points occur where the inequality becomes undefined, which is when the denominators are zero. Solve \( x - 1 = 0 \) and \( x = 0 \). This gives us two critical points: \( x = 1 \) and \( x = 0 \).

Step 2: Rewrite the Inequality

Move all terms to one side of the inequality: \( \frac{3}{x-1} - \frac{4}{x} - 1 \geq 0 \). Find a common denominator, \( x(x-1) \), which yields \( \frac{3x - 4(x-1) - x(x-1)}{x(x-1)} \geq 0 \).

Step 3: Simplify the Numerator

Simplify the numerator: \( 3x - 4x + 4 - x^2 + x \). Combine like terms to get \( -x^2 + 4 \). Thus, the inequality is \( \frac{-x^2 + 4}{x(x-1)} \geq 0 \).

Step 4: Identify Key Intervals

Identify intervals between and beyond the critical points \( x = 0 \) and \( x = 1 \): \((-\infty, 0)\), \((0, 1)\), and \((1, \infty)\).

Step 5: Test Each Interval

Choose a test point from each interval to determine if the interval satisfies the inequality: 1. For \((-\infty, 0)\), choose \(x = -1\).2. For \((0,1)\), choose \(x = 0.5\).3. For \((1, \infty)\), choose \(x = 2\).

Step 6: Evaluate and Solve

Compute the inequality for each test point:1. \(x = -1\): \( \frac{-(-1)^2 + 4}{-1 \cdot (-1-1)} = \frac{-1+4}{2} > 0\), true.2. \(x = 0.5\): \( \frac{-(0.5)^2 + 4}{0.5(-0.5)} \approx 0 < 0\), false.3. \( x = 2 \): \( \frac{-(2)^2 +4}{2(2-1)} = 0\), satisfies equality.

Step 7: Confirm Solution and Express in Interval Notation

The inequality holds true for \((-\infty, 0)\) and \([1, \infty)\). Confirm endpoints: \(x = 1\) does not satisfy the inequality due to undefined behavior, so use:\((-\infty, 0) \cup (1, \infty) \).

Key Concepts

Interval Notation

When solving nonlinear inequalities, expressing the solution using interval notation makes understanding the solution easier and more visual. Think of interval notation as a way to represent sets of numbers on a number line.

We often express an entire spectrum of solutions as ranges from a starting point to an endpoint. These ranges might include the endpoints themselves or exclude them, depending on the problem.

• An open interval, written as \(a, b\), means all numbers between \(a\) and \(b\), not including \(a\) or \(b\).
• A closed interval, written as \[a, b\]], means all numbers from \(a\) to \(b\), including \(a\) and \(b\).
• Mixed intervals, like \(a, b\]] or \[a, b\), include one endpoint but not the other.

Understanding this makes graphing solutions and conveying them clearly straightforward, as seen in our example where the solution is expressed as \((-\infty, 0) \cup (1, \infty)\).

Critical Points

In non-linear inequalities, critical points are crucial as they are where the polynomial equals zero or becomes undefined. Discovering these points helps in breaking the number line into sections or intervals that can be separately evaluated.

To find the critical points, look where the function is undefined or zero.

• In our example, the critical points are found by setting the denominators of each term to zero: \((x-1) = 0\) and \(x = 0\).
• The critical points are \(x=1\) and \(x=0\).

These points help us choose test points within each interval, ensuring we adequately determine where the inequality holds or doesn't hold.

Test Intervals

Test intervals are sections on the number line created between critical points. By selecting test points within each interval, we determine whether an interval satisfies the inequality.

After establishing critical points, you have sections like \((-\infty, 0)\), (0, 1), \(1, \infty)\).

• Choose a simple test point within each interval.
• Plug these test points into the simplified inequality.

Is the inequality true or false at these points? Different intervals may behave differently.
For example, in our inequality:

• For \((-\infty, 0)\), test point \(x = -1\) results in a true condition.
• For \(0, 1\), test point \(x = 0.5\) does not satisfy the inequality.
• For \(1, \infty)\), test point \(x = 2\) is valid, showing it meets the requirement.

These evaluations help us craft a precise solution set, expressed in interval notation as \((-\infty, 0) \cup (1, \infty)\), eliminating guesswork.