Problem 88
Question
Solve each rational inequality. Write each solution set in interval notation.4 $$\frac{9 x-8}{4 x^{2}+25}<0$4
Step-by-Step Solution
Verified Answer
\((-\infty, \frac{8}{9})\)
1Step 1: Identify Critical Points
Find the values of x that make the numerator and the denominator zero. This helps in identifying the intervals to test. The numerator is zero when \(9x - 8 = 0\), leading to \(x = \frac{8}{9}\). The denominator is always positive since \(4x^2 + 25 > 0\) for all real x.
2Step 2: Determine the Sign of the Expression
Evaluate the sign of \(\frac{9x-8}{4x^2+25}\) in the identified intervals. Since the denominator is always positive, the sign of the expression depends on the numerator. For \(x <\frac{8}{9}\), test a point, e.g., \(x = 0\). For \(x > \frac{8}{9}\), test a point, e.g., \(x = 1\). Substitute these values back into the inequality:
3Step 3: Test Interval for x < 8/9
Substituting \(x = 0\) into the numerator: \(9(0) - 8 = -8\), which is negative. Since the denominator is positive, the overall expression is negative in this interval.
4Step 4: Test Interval for x > 8/9
Substituting \(x = 1\) into the numerator: \(9(1) - 8 = 1\), which is positive. Since the denominator is positive, the overall expression is positive in this interval.
5Step 5: Write the Solution Set
From the tests, the expression \(\frac{9x-8}{4x^2+25} < 0\) is true when \(x < \frac{8}{9}\). Write the solution set in interval notation: \((-\infty, \frac{8}{9})\).
Key Concepts
Critical PointsInterval NotationSign Analysis
Critical Points
Critical points are essential in solving rational inequalities. They are the values of x where the numerator or the denominator of a rational expression equals zero.
Finding these values helps in breaking down the number line into intervals that you can test to determine where the inequality holds.
For the given exercise, the numerator is zero when \(9x - 8 = 0\), hence \(x = \frac{8}{9}\). This is a critical point as it makes the numerator zero.
The denominator, \(4x^2 + 25\), never equals zero since it's always positive for any real x. Therefore, it doesn't contribute any additional critical points but confirms that we won't divide by zero, avoiding undefined expressions.
In summary, for the inequality \(\frac{9x-8}{4x^2+25}<0\), the critical point is \(x = \frac{8}{9}\). Identifying this point sets up the stage for further steps like interval testing.
Finding these values helps in breaking down the number line into intervals that you can test to determine where the inequality holds.
For the given exercise, the numerator is zero when \(9x - 8 = 0\), hence \(x = \frac{8}{9}\). This is a critical point as it makes the numerator zero.
The denominator, \(4x^2 + 25\), never equals zero since it's always positive for any real x. Therefore, it doesn't contribute any additional critical points but confirms that we won't divide by zero, avoiding undefined expressions.
In summary, for the inequality \(\frac{9x-8}{4x^2+25}<0\), the critical point is \(x = \frac{8}{9}\). Identifying this point sets up the stage for further steps like interval testing.
Interval Notation
Once you identify critical points, you can break the real number line into intervals. Interval notation helps describe the solution set of inequalities concisely and visually.
For example, the critical point \(x = \frac{8}{9}\) splits the number line into two intervals: \((-\infty, \frac{8}{9})\) and \((\frac{8}{9}, \infty)\).
Interval notation allows you to clearly specify where an inequality holds or doesn't hold. In the solution of our exercise: the interval \((-\infty, \frac{8}{9})\) indicates that values of x in this range satisfy the rational inequality \(\frac{9x-8}{4x^2+25}<0\).
Using parentheses \(()\) indicates that endpoints are not included in the interval. Brackets \([]\) would be used if endpoints were to be included. However, in our case, the critical point \(x = \frac{8}{9}\) doesn't satisfy \(\frac{9x-8}{4x^2+25}<0\) and thus is not included in the interval.
For example, the critical point \(x = \frac{8}{9}\) splits the number line into two intervals: \((-\infty, \frac{8}{9})\) and \((\frac{8}{9}, \infty)\).
Interval notation allows you to clearly specify where an inequality holds or doesn't hold. In the solution of our exercise: the interval \((-\infty, \frac{8}{9})\) indicates that values of x in this range satisfy the rational inequality \(\frac{9x-8}{4x^2+25}<0\).
Using parentheses \(()\) indicates that endpoints are not included in the interval. Brackets \([]\) would be used if endpoints were to be included. However, in our case, the critical point \(x = \frac{8}{9}\) doesn't satisfy \(\frac{9x-8}{4x^2+25}<0\) and thus is not included in the interval.
Sign Analysis
Sign analysis is the process of determining the sign (positive or negative) of an expression within the intervals created by critical points.
By testing a value from each interval, you can determine where the rational expression is negative or positive. This is crucial for understanding where the inequality holds.
In the given exercise, we tested \(x = 0\) in the interval \(x < \frac{8}{9}\) and found that \(\frac{9(0)-8}{4(0)^2+25}=-\frac{8}{25}\), which is negative.
For the interval \(x > \frac{8}{9}\), testing \(x = 1\) gives \(\frac{9(1)-8}{4(1)^2+25}=\frac{1}{29}\), which is positive.
Therefore, the interval \((-\infty, \frac{8}{9})\) is part of the solution, while \((\frac{8}{9}, \infty)\) is not. This is how sign analysis confirms whether an interval should be included in the solution set for the inequality.
By testing a value from each interval, you can determine where the rational expression is negative or positive. This is crucial for understanding where the inequality holds.
In the given exercise, we tested \(x = 0\) in the interval \(x < \frac{8}{9}\) and found that \(\frac{9(0)-8}{4(0)^2+25}=-\frac{8}{25}\), which is negative.
For the interval \(x > \frac{8}{9}\), testing \(x = 1\) gives \(\frac{9(1)-8}{4(1)^2+25}=\frac{1}{29}\), which is positive.
Therefore, the interval \((-\infty, \frac{8}{9})\) is part of the solution, while \((\frac{8}{9}, \infty)\) is not. This is how sign analysis confirms whether an interval should be included in the solution set for the inequality.
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