Advanced Curve Sketching and Second Derivative Analysis

Inconvenient Values in Curve Sketching: The instructor notes that in curve sketching problems, stationary points and points of inflection often result in "inconvenient numbers."
Trade-off in Problem Design: When creating these questions, if one point (like a stationary point) is made into a "nice" number (e.g., $1$ or $2$ ), the other points (like inflection points) often become mathematically complex or "haywire."
Second Derivative Critical Points: For the function being discussed, the critical values for the second derivative are: - $x = -\sqrt{e^{5/3}}$ - $x = 0$ - $x = +\sqrt{e^{5/3}}$
Importance of Zero on the Number Line: Even if the function is undefined at $x = 0$ , this value must be placed on the second derivative number line to accurately determine intervals of concavity.

Second Derivative Denominator: In this specific problem, the denominator of the second derivative is always positive. This contrast with the first derivative, where the denominator's sign might change, necessitates careful evaluation.
Testing Intervals for Concavity: - Interval $x < -\sqrt{e^{5/3}}$ : Choosing a test point like $x = -10$ . Plugging this into the numerator $3 \times \ln(x^2) - 5$ . Since $(-10)^2 = 100$ , the logarithmic term is large and positive, resulting in a positive numerator and a positive denominator. Thus, the graph is concave up. - Interval between $-\sqrt{e^{5/3}}$ and $0$ : Choosing a test point like $x = -1$ . Plugging this into the numerator: $(-1)^2 = 1$ , and $\ln(1) = 0$ . The term $0 - 5$ is negative. Resulting in a negative value (Concave Down). - Interval between $0$ and $\sqrt{e^{5/3}}$ : Choosing a test point like $x = 1$ . The numerator remains negative ( $\ln(1) - 5$ ) and the denominator is positive. The concavity stays concave down.
Caveat on Sign Alternation: Concavity does not automatically alternate signs at every critical point. In this instance, the graph maintains concavity across the asymptote at $x = 0$ .
Summary of Concavity Changes: - From left to right: Concave up, then concave down, remains concave down past zero, then changes back to concave up after the positive root.

Undefined Points: Step 4 of the lecture slides applies when a function is undefined. In this case, the function is not defined at $x = 0$ .
Limits as $x \to 0$ : - Formula: $f(x) = \frac{\ln(x^2)}{x^2}$ - As $x \to 0$ , $\ln(x^2) \to -\infty$ . - The denominator $x^2$ approaches a very small positive number. - A very large negative number divided by a very small positive number result in negative infinity: $\lim_{x \to 0} \frac{\ln(x^2)}{x^2} = -\infty$ . - This holds for both the left-hand approach ( $0^-$ ) and right-hand approach ( $0^+$ ) because the $x^2$ in the denominator guarantees a positive divisor.
Limits as $x \to \pm\infty$ : - This results in an $\frac{\infty}{\infty}$ indeterminate form, allowing the use of L'Hôpital's Rule ( $L'H$ ). - Differentiating the numerator $\ln(x^2)$ gives $\frac{2x}{x^2}$ , and differentiating the denominator $x^2$ gives $2x$ . - Simplification: $\frac{2/x}{2x} = \frac{1}{x^2}$ . - As $x \to \infty$ or $x \to -\infty$ , the value $\frac{1}{x^2} \to 0$ .
Asymptote Behavior: Because $1/x^2$ is positive, the graph approaches the x-axis ( $y = 0$ ) from slightly above.

Key Coordinates for Plotting: - Intercepts: $x = -1$ and $x = 1$ . - Stationary Points: $\pm\sqrt{e}$ . - Inflection Points: $\pm\sqrt{e^{5/3}}$ .
Step-by-Step Sketching: 1. Left Side ( $x < 0$ ): Start near the x-axis ( $y=0$ ) at negative infinity. The graph is increasing and concave up until it reaches the inflection point at $-\sqrt{e^{5/3}}$ . 2. Transition: At $-\sqrt{e^{5/3}}$ , it changes to concave down but continues to increase until it hits a horizontal stationary point at $-\sqrt{e}$ . 3. Descent: After the maximum at $-\sqrt{e}$ , the graph slopes down, passes through the x-intercept at $-1$ , and plunges toward negative infinity at the vertical asymptote $x = 0$ . 4. Right Side ( $x > 0$ ): The graph emerges from negative infinity at $x = 0$ . It is sloping up and concave down. 5. Peak: It passes through the x-intercept at $1$ and reaches a maximum at $\sqrt{e}$ . 6. Completion: After $\sqrt{e}$ , it slopes down. At the inflection point $\sqrt{e^{5/3}}$ , it becomes concave up and asymptotically approaches the x-axis ( $y=0$ ).

Class Test 2 vs. Final Exam: - For Class Test 2, clarity on the number line and indicating concave effects is usually sufficient for full marks. - For the Final Exam, students should explicitly write out intervals for concavity (e.g., "Concave up for $x < -\sqrt{e^{5/3}}$ and $x > \sqrt{e^{5/3}}$ ") to ensure examiners do not deduct marks.
Likely Test Topics: - Differentiation formulas requiring the Chain Rule. - Implicit Differentiation where $x$ and $y$ are jumbled together. - Newton's Method (ensure calculator proficiency). - Optimization Problems: Finding maximums/minimums (e.g., maximizing profit). - Curve Sketching: Differentiating, finding intercepts, and limits.
Trigonometry in Exams: While trigonometric functions (like $\tan(x)$ or $\arcsin(x)$ ) might appear inside chain rule or implicit differentiation problems, they are unlikely to be the primary focus of complex curve sketching or identity-heavy questions.

Continuity of the Function: - Question: Is the formula/graph continuous or discontinuous? - Answer: The function is continuous on its domain. - Explanation: Continuity is only defined where the function exists. While there is a break at $x = 0$ , the function is not defined there; therefore, it is continuous everywhere it is defined. - Comparison Example: A graph with an open circle that is filled elsewhere is discontinuous. A graph that simply does not exist at a point (like this one at zero) is continuous on its domain.
Limits vs. Continuity: The limit as $x \to 0$ does not exist (it is infinity), but this does not affect the continuity of the function within its domain because $x = 0$ is excluded from that domain.