Lecture 4.5/4.7

Classifier Type: A type of classifier involving slant asymptotes.
- Affinity: Classifiers based on directionality (positive or negative).
Slant Asymptotes:
- Do not approach a limit or minimum value, rather grow to infinity or negative infinity.
- Direction is defined by the equation of a straight line:
  $y = nx + b$
- Mathematical Meaning: The curve approaches the line where the difference between the function and the line approaches zero as $x$ approaches infinity.

Rational Functions:
- Asymptotes appear when the degree of the numerator is one greater than the degree of the denominator.
- Example Calculation: Given the rational function
  $f(x) = \frac{x^3}{x^2 + 1}$
- Degree of the numerator (3) is 1 greater than the degree of the denominator (2).
- No Vertical Asymptotes: As $x^2 + 1$ is never equal to zero.

Behavior at Infinity:
- As $x o +\infty$ , $f(x) o +\infty$ .
- As $x o -\infty$ , $f(x) o -\infty$ .
Domain:
- For the given function
  $f(x) = \frac{x^3}{x^2 + 1}$
- The domain is all real numbers, $(-\infty, +\infty)$ .

Intercepts:
- Crosses the y-axis when $x = 0$ , yielding $f(0) = 0$ .
- Crosses the x-axis also at $f(x) = 0$ , resulting in the single intercept at the origin (0,0).
Symmetry:
- The function shows odd symmetry:
- Since the numerator ( $x^3$ ) is odd and the denominator ( $x^2 + 1$ ) is even, $f(-x) = -f(x)$ .
- Function is symmetric about the origin: Reflects across the origin, thus simplifying graph plotting.

Vertical Asymptotes:
- As $x^2 + 1$ is never equal to zero, there are no vertical asymptotes.
Long Division for Slant Asymptotes:
- Perform long division on $\frac{x^3}{x^2 + 1}$ to find slant asymptotes:
1. Divide $x^3$ by $x^2 + 1$ .
2. Result in a linear expression with a remainder. Use this to determine the asymptote.

First Derivative:
- Important for determining intervals of increase and decrease.
- Use the Quotient Rule for differentiation:
- If $f(x) = \frac{p(x)}{q(x)}$ , then
  $f'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{(q(x))^2}$ .
Critical Points:
- Found by setting the first derivative to zero to solve for $x$ .

Second Derivative Test:
- Provides information about concavity and inflection points.
- Solving for points where the second derivative equals zero helps determine whether the function is concave up or down.

Word Problem Setup:
- Start with the approach word “Let” to define variables based on the problem.
- Identify the unknowns and write equations to represent constraints.
Example:
- Fencing problem involving a rectangular area bordered by a river, maximizing the area based on given constraints (e.g., number of sides constrained by fencing).
- Define variables for width $x$ and length $y$ , leading to the equations: $x + y = 2400$ for optimization of area: $A = x \cdot y$ .

Maximization Approach:
- Use derivatives to find local maxima and minima.
- Absolute extrema occur at critical points and endpoints.
- Substitute values for area and constraints to find optimal configurations.

Cost and Revenue Functions:
- Cost function $C(x)$ describes the cost of producing $x$ units.
- Revenue is described by the revenue function $R(x) = x \cdot P$ , where $P$ is price per unit.
- Profit is computed as:
  $Profit = Revenue - Cost$ .
Maximizing Profit:
- Differentiate the profit function to find critical points.
- Assess behavior of functions through marginal analysis (i.e., marginal cost, marginal revenue).

Monitor Sales Case Study:
- Given base price and demand increases with rebates, define demand function considering price adjustments. Derive revenue function from demand.
- Set derivatives to zero to find revenue-maximizing quantity and price.
Conclusion: Always analyze endpoints and found critical points to ensure proper identification of extrema and be aware that changes in pricing directly impact demand and subsequent revenue.