Lab 8: Solving Various Types of Inequalities

Students are urged to read the material prior to attending the lab.
Emphasis on the importance of preparation to avoid slowing down the group.
Overview of the relevance of derivatives in calculus, specifically how they provide information about a function and its graph.
The lab focuses on the necessary algebra and trigonometry skills to extract information from inequalities and equations.

By solving equations, accompanying inequalities can also be tackled effectively.

Step 1: Solve the Equation ( x^2 - 5x - 24 = 0 )
- Use factoring: ( (x - 8)(x + 3) = 0 )
- Solutions: ( x = 8 ) or ( x = -3 )
Step 2: Determine Positivity of the Expression
- Expression is zero at ( x = 8 ) and ( x = -3 )
- Find intervals and test points to see where the expression is greater than zero:

Number line divided into three sections based on critical points ( -3 ) and ( 8 ):
- Interval 1: (-∞, -3)
- Interval 2: (-3, 8)
- Interval 3: (8, ∞)
Test Points:
1. For ( x = -4 ):
- Evaluating: ( (x - 8)(x + 3) = (-12)(-1) = 12 )
- Result: positive (place a positive sign above this interval)
1. For ( x = 0 ):
- Evaluating: ( (0 - 8)(0 + 3) = (-8)(3) = -24 )
- Result: negative (place a negative sign above this interval)
1. For ( x = 9 ):
- Evaluating: ( (9 - 8)(9 + 3) = (1)(12) = 12 )
- Result: positive (place a positive sign above this interval)
Conclusion of Inequality Solution:
- The solution set is ( \, {x | x \leq -3 \text{ or } x > 8} )
- In interval notation, this is represented as ( (-\infty, -3] \cup [8, \infty) )

Interval	Test Value	(x - 8)	(x + 3)	(x - 8)(x + 3)	Conclusion
(-∞, -3)	-4	-12 (-)	-1 (−)	12 (+)	positive
(-3, 8)	0	-8 (-)	3 (+)	-24 (-)	negative
(8, ∞)	9	1 (+)	12 (+)	12 (+)	positive

Both methods yield the same solution set.
Positive or negative clarity is enough instead of numeric values from test points.
Visualization is key; understanding that the graph of a parabola opens upward is crucial.
The expression is above the x-axis (positive) at the extremes of the x-values, approaching ( -\infty ) or ( +\infty ).

Identification of behavior of functions in terms of increasing or decreasing based on the first derivative.
Students should review simplifying derivatives prior to moving on to more complex scenarios.

Critical Numbers: Defined as numbers ( c ) in the domain of ( f ) where ( f'(c) = 0 ) or ( f'(c) ) does not exist:
- Found critical values: ( x = 4 ), ( x = 0 ), and ( x = 6 )

Various test values determined the sign of the derivative in different intervals:
1. For ( x = -1 ): Derivative negative
2. For ( x = 1 ): Derivative positive
3. For ( x = 5 ): Derivative negative
4. For ( x = 7 ): Derivative negative
Conclusion of Example 8.2:
- ( f'(x) > 0 ) on ( (0, 4) )
- ( f'(x) < 0 ) on ( (-\infty, 0) \cup (4, 6) \cup (6, \infty) )

Derivative Calculation:
- Utilized the product rule to find ( f'(x) )
- Critical values determined: ( 0 ) and ( \frac{1}{2} )
- It was noted that the ( e^x ) portion does not provide critical numbers since it is always positive.

Determined derivative signs based on test points in relevant intervals:
- Positive Interval: ( (0, \frac{1}{2}) )
- Negative Interval: ( (\frac{1}{2}, \infty) )

Given ( f(0) = 0 + ext{cos}(0) )
Derivative found: ( f'(0) = 1 - ext{sin}(0) )
Critical numbers are determined from the derivatives based on the interval restrictions.

Positive conclusions made at intervals based on selected test values, yielding understanding of function behavior.
It was noted that since (-1 \leq ext{sin}(x) \leq 1), then ( f'(0) = 1 - ext{sin}(x) \geq 0) for all applicable values.