Answer :
### Part (a): Algebraic Solution
Given the inequality:
[tex]\[ \frac{-6t}{t-2} < \frac{-30}{t-2} \][/tex]
We will solve this step by step:
1. Combine the terms into a single fraction:
[tex]\[ \frac{-6t}{t-2} - \frac{-30}{t-2} < 0 \][/tex]
Since the two fractions have the same denominator, we can combine the numerators:
[tex]\[ \frac{-6t + 30}{t-2} < 0 \][/tex]
2. Factor the numerator:
[tex]\[ \frac{-6(t-5)}{t-2} < 0 \][/tex]
3. Identify the critical points:
The critical points are the values of [tex]\( t \)[/tex] that make either the numerator or the denominator equal to zero:
[tex]\[ -6(t-5) = 0 \quad \Rightarrow \quad t = 5 \][/tex]
[tex]\[ t-2 = 0 \quad \Rightarrow \quad t = 2 \][/tex]
4. Determine the sign of the expression in the intervals divided by these critical points:
The critical points [tex]\( t = 2 \)[/tex] and [tex]\( t = 5 \)[/tex] divide the real number line into three intervals:
- [tex]\((-∞, 2)\)[/tex]
- [tex]\((2, 5)\)[/tex]
- [tex]\((5, ∞)\)[/tex]
We need to test the sign of [tex]\(\frac{-6(t-5)}{t-2}\)[/tex] within each of these intervals.
- Interval [tex]\((-∞, 2)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 1 \)[/tex]):
[tex]\[ \frac{-6(1-5)}{1-2} = \frac{-6(-4)}{-1} = \frac{24}{-1} = -24 \quad (\text{Negative}) \][/tex]
- Interval [tex]\((2, 5)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 3.5 \)[/tex]):
[tex]\[ \frac{-6(3.5-5)}{3.5-2} = \frac{-6(-1.5)}{1.5} = \frac{9}{1.5} = 6 \quad (\text{Positive}) \][/tex]
- Interval [tex]\((5, ∞)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 6 \)[/tex]):
[tex]\[ \frac{-6(6-5)}{6-2} = \frac{-6(1)}{4} = \frac{-6}{4} = -1.5 \quad (\text{Negative}) \][/tex]
5. Combine the results:
We found that the expression is negative in the intervals [tex]\((-∞, 2)\)[/tex] and [tex]\((5, ∞)\)[/tex]. Since we want our inequality to be less than zero, these intervals are our solution. However, we must exclude the points where the expression is undefined (i.e., [tex]\( t = 2 \)[/tex]) and where the inequality is exactly zero (i.e., [tex]\( t = 5 \)[/tex]).
Thus, the solution set is:
[tex]\[ (-∞, 2) \cup (5, ∞) \][/tex]
### Part (b): Geometric Solution
To support our algebraic solution, we can use a geometric approach by considering the function [tex]\( f(t) = \frac{-6(t-5)}{t-2} \)[/tex].
1. Sketch the function:
The function [tex]\( f(t) = \frac{-6(t-5)}{t-2} \)[/tex] has the following key features:
- Vertical asymptote at [tex]\( t = 2 \)[/tex] (where the denominator is zero).
- Zero at [tex]\( t = 5 \)[/tex] (where the numerator is zero).
2. Analysis of asymptotes and sign changes:
- For [tex]\( t < 2 \)[/tex]: As [tex]\( t \)[/tex] approaches 2 from the left, the denominator [tex]\( t-2 \)[/tex] is negative. The numerator [tex]\( -6(t-5) \)[/tex] is positive because [tex]\( t-5 \)[/tex] is negative and the expression will be negative. This supports the algebraic analysis where the expression is negative in [tex]\( (-∞, 2) \)[/tex].
- For [tex]\( t \)[/tex] between 2 and 5:
- The numerator [tex]\( -6(t-5) \)[/tex] is positive because [tex]\( t-5 \)[/tex] is negative.
- The denominator [tex]\( t-2 \)[/tex] is positive.
- Therefore, the fraction is positive. This matches our algebraic solution.
- For [tex]\( t > 5 \)[/tex]:
- The numerator [tex]\( -6(t-5) \)[/tex] is negative because [tex]\( t-5 \)[/tex] is positive.
- The denominator [tex]\( t-2 \)[/tex] is positive.
- Therefore, the fraction is negative. This corresponds to the interval [tex]\( (5, ∞) \)[/tex] where the expression is negative.
This geometric analysis confirms that the inequality [tex]\(\frac{-6(t-5)}{t-2} < 0\)[/tex] holds true for the intervals [tex]\( (-∞, 2) \cup (5, ∞) \)[/tex], supporting our algebraic solution.
Given the inequality:
[tex]\[ \frac{-6t}{t-2} < \frac{-30}{t-2} \][/tex]
We will solve this step by step:
1. Combine the terms into a single fraction:
[tex]\[ \frac{-6t}{t-2} - \frac{-30}{t-2} < 0 \][/tex]
Since the two fractions have the same denominator, we can combine the numerators:
[tex]\[ \frac{-6t + 30}{t-2} < 0 \][/tex]
2. Factor the numerator:
[tex]\[ \frac{-6(t-5)}{t-2} < 0 \][/tex]
3. Identify the critical points:
The critical points are the values of [tex]\( t \)[/tex] that make either the numerator or the denominator equal to zero:
[tex]\[ -6(t-5) = 0 \quad \Rightarrow \quad t = 5 \][/tex]
[tex]\[ t-2 = 0 \quad \Rightarrow \quad t = 2 \][/tex]
4. Determine the sign of the expression in the intervals divided by these critical points:
The critical points [tex]\( t = 2 \)[/tex] and [tex]\( t = 5 \)[/tex] divide the real number line into three intervals:
- [tex]\((-∞, 2)\)[/tex]
- [tex]\((2, 5)\)[/tex]
- [tex]\((5, ∞)\)[/tex]
We need to test the sign of [tex]\(\frac{-6(t-5)}{t-2}\)[/tex] within each of these intervals.
- Interval [tex]\((-∞, 2)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 1 \)[/tex]):
[tex]\[ \frac{-6(1-5)}{1-2} = \frac{-6(-4)}{-1} = \frac{24}{-1} = -24 \quad (\text{Negative}) \][/tex]
- Interval [tex]\((2, 5)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 3.5 \)[/tex]):
[tex]\[ \frac{-6(3.5-5)}{3.5-2} = \frac{-6(-1.5)}{1.5} = \frac{9}{1.5} = 6 \quad (\text{Positive}) \][/tex]
- Interval [tex]\((5, ∞)\)[/tex]:
Choose a test point in this interval (e.g., [tex]\( t = 6 \)[/tex]):
[tex]\[ \frac{-6(6-5)}{6-2} = \frac{-6(1)}{4} = \frac{-6}{4} = -1.5 \quad (\text{Negative}) \][/tex]
5. Combine the results:
We found that the expression is negative in the intervals [tex]\((-∞, 2)\)[/tex] and [tex]\((5, ∞)\)[/tex]. Since we want our inequality to be less than zero, these intervals are our solution. However, we must exclude the points where the expression is undefined (i.e., [tex]\( t = 2 \)[/tex]) and where the inequality is exactly zero (i.e., [tex]\( t = 5 \)[/tex]).
Thus, the solution set is:
[tex]\[ (-∞, 2) \cup (5, ∞) \][/tex]
### Part (b): Geometric Solution
To support our algebraic solution, we can use a geometric approach by considering the function [tex]\( f(t) = \frac{-6(t-5)}{t-2} \)[/tex].
1. Sketch the function:
The function [tex]\( f(t) = \frac{-6(t-5)}{t-2} \)[/tex] has the following key features:
- Vertical asymptote at [tex]\( t = 2 \)[/tex] (where the denominator is zero).
- Zero at [tex]\( t = 5 \)[/tex] (where the numerator is zero).
2. Analysis of asymptotes and sign changes:
- For [tex]\( t < 2 \)[/tex]: As [tex]\( t \)[/tex] approaches 2 from the left, the denominator [tex]\( t-2 \)[/tex] is negative. The numerator [tex]\( -6(t-5) \)[/tex] is positive because [tex]\( t-5 \)[/tex] is negative and the expression will be negative. This supports the algebraic analysis where the expression is negative in [tex]\( (-∞, 2) \)[/tex].
- For [tex]\( t \)[/tex] between 2 and 5:
- The numerator [tex]\( -6(t-5) \)[/tex] is positive because [tex]\( t-5 \)[/tex] is negative.
- The denominator [tex]\( t-2 \)[/tex] is positive.
- Therefore, the fraction is positive. This matches our algebraic solution.
- For [tex]\( t > 5 \)[/tex]:
- The numerator [tex]\( -6(t-5) \)[/tex] is negative because [tex]\( t-5 \)[/tex] is positive.
- The denominator [tex]\( t-2 \)[/tex] is positive.
- Therefore, the fraction is negative. This corresponds to the interval [tex]\( (5, ∞) \)[/tex] where the expression is negative.
This geometric analysis confirms that the inequality [tex]\(\frac{-6(t-5)}{t-2} < 0\)[/tex] holds true for the intervals [tex]\( (-∞, 2) \cup (5, ∞) \)[/tex], supporting our algebraic solution.
