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Questions and Answers
When solving a rational inequality, what is the significance of the values obtained by setting the numerator and denominator to zero?
When solving a rational inequality, what is the significance of the values obtained by setting the numerator and denominator to zero?
- They determine the end behavior of the function as x approaches infinity.
- They define the critical points used to determine intervals on a number line. (correct)
- They indicate where the function is increasing or decreasing.
- They represent the y-intercepts of the rational function.
Why is it necessary to rearrange a rational inequality so that one side is zero before solving it?
Why is it necessary to rearrange a rational inequality so that one side is zero before solving it?
- To accurately determine the intervals where the inequality holds true by comparing to zero. (correct)
- To simplify the algebraic manipulation.
- To eliminate the denominator and solve for x directly.
- To ensure that the critical points are easily identifiable.
In the inequality $\frac{(x - 3)}{(x + 2)} ≥ 0$, why does x = -2 have an open circle on the number line?
In the inequality $\frac{(x - 3)}{(x + 2)} ≥ 0$, why does x = -2 have an open circle on the number line?
- Because the inequality includes 'equal to.'
- Because x = -2 makes the denominator zero, resulting in an undefined expression. (correct)
- Because x = -2 is a negative number.
- Because x = -2 makes the numerator zero.
Given the inequality $\frac{(x - 4)(x + 1)}{(x - 3)} < 0$, how do you determine the intervals where the inequality holds true?
Given the inequality $\frac{(x - 4)(x + 1)}{(x - 3)} < 0$, how do you determine the intervals where the inequality holds true?
For the inequality $\frac{(x + 2)}{(x - 1)} ≤ 3$, what is the first step after rearranging the inequality to have zero on one side?
For the inequality $\frac{(x + 2)}{(x - 1)} ≤ 3$, what is the first step after rearranging the inequality to have zero on one side?
What does the union symbol '∪' represent when expressing the solution to a rational inequality in interval notation?
What does the union symbol '∪' represent when expressing the solution to a rational inequality in interval notation?
In solving rational inequalities, when do you use a closed circle on the number line for a critical point?
In solving rational inequalities, when do you use a closed circle on the number line for a critical point?
Consider the rational inequality $\frac{x+a}{x-b} < 0$. What must be true about the values of $x$ in the solution?
Consider the rational inequality $\frac{x+a}{x-b} < 0$. What must be true about the values of $x$ in the solution?
What is the solution set for the inequality $\frac{1}{x} > 1$?
What is the solution set for the inequality $\frac{1}{x} > 1$?
For what values of $x$ is the inequality $\frac{x^2 - 1}{x - 2} ≥ 0$ satisfied?
For what values of $x$ is the inequality $\frac{x^2 - 1}{x - 2} ≥ 0$ satisfied?
Flashcards
Solving Rational Inequalities
Solving Rational Inequalities
Values of x that satisfy the inequality.
Critical Points
Critical Points
Points where the numerator or denominator of a rational expression equals zero.
Number Line in Inequalities
Number Line in Inequalities
A visual tool to determine the intervals where the inequality is true.
Testing Intervals
Testing Intervals
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Interval Notation
Interval Notation
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Rearranging Inequalities
Rearranging Inequalities
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Common Denominator
Common Denominator
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Denominator Restrictions
Denominator Restrictions
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Multiplicity of a Zero
Multiplicity of a Zero
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Study Notes
Solving Rational Inequalities
- The goal is to find the values of x for which the inequality holds true.
- A number line helps visualize intervals determined by critical points.
- Critical points are found by setting the numerator and the denominator of the rational expression equal to zero.
Example 1: (x - 3) / (x + 2) ≥ 0
- Setting the numerator to zero: x - 3 = 0, which means x = 3.
- Setting the denominator to zero: x + 2 = 0, which means x = -2.
- On the number line, x = -2 is an open circle because the denominator cannot equal zero, and x = 3 is a closed circle because the inequality includes "equal to".
- Test values from each region (less than -2, between -2 and 3, greater than 3) in the original inequality to determine the sign of the result.
- If x = 4, (4 - 3) / (4 + 2) yields positive/positive = positive.
- If x = 0, (0 - 3) / (0 + 2) yields negative/positive = negative.
- If x = -3, (-3 - 3) / (-3 + 2) yields negative/negative = positive.
- Shade the regions where the results are positive (greater than or equal to zero).
- Interval notation solution: (-∞, -2) ∪ [3, ∞).
- Inequality solution: x < -2 or x ≥ 3.
Example 2: (x - 4)(x + 1) / (x - 3) < 0
- Points of interest: x = 4, x = -1, and x = 3.
- Because the inequality is strictly less than zero, all points of interest are open circles.
- Test values from each region (less than -1, between -1 and 3, between 3 and 4, greater than 4) in the original inequality to figure out the sign of the result.
- Signs alternate if the multiplicity of each zero is one.
- Interval notation solution: (-∞, -1) ∪ (3, 4).
- Inequality solution: x < -1 or 3 < x < 4.
Example 3: (x + 2) / (x - 1) ≤ 3
- Because the inequality is not compared to zero, rearrange it to have zero on one side by subtracting 3 from both sides: (x + 2) / (x - 1) - 3 ≤ 0.
- Find a common denominator by multiplying -3 by (x - 1) / (x - 1).
- Combine terms into a single fraction: (x + 2 - 3(x - 1)) / (x - 1) ≤ 0.
- Simplify the expression: (-2x + 5) / (x - 1) ≤ 0.
- The denominator cannot be zero, so x ≠ 1, represented by an open circle at x = 1.
- Setting the numerator to zero: -2x + 5 = 0, so x = 5/2 = 2.5, represented by a closed circle at x = 2.5.
- Test values from each region (less than 1, between 1 and 2.5, greater than 2.5) to determine the sign.
- If x = 3, then (-2 * 3 + 5) / (3 - 1) is negative.
- If x = 2, then (-2 * 2 + 5) / (2 - 1) is positive.
- If x = 0, then (-2 * 0 + 5) / (0 - 1) is negative.
- Shade the regions less than or equal to zero, so shade the negative regions.
- Interval notation solution: (-∞, 1) ∪ [5/2, ∞).
- Inequality solution: x < 1 or x ≥ 5/2.
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