Session 13 - Critical Points and Extrema | Tony's Intertwingled Memex

📅 Micro-Lecture

1. Optimization Design & The Geometry of Critical Points

Mathematical Foundation: An extreme value of a function represents an optimum: the greatest (maximum) or smallest (minimum) value achievable over a region. Calculus identifies candidate points by inspecting rates of change. A critical number of a function $f$ is any value $c$ in its domain ( $c \in \text{dom}(f)$ ) where either the tangent is horizontal ( $f'(c) = 0$ ) or the curve experiences a sharp corner, cusp, or vertical tangent ( $f'(c)$ does not exist).
Engineering Context: In civil, mechanical, and electrical engineering, locating critical points is the core engine of optimization. Structural engineers calculate zero-shear locations ( $\frac{dM}{dx} = 0$ ) to locate maximum beam bending moments, aerospace engineers pinpoint peak dynamic pressure ( $q$ -Max) to size airframes, and electrical engineers locate peak transistor junction temperatures to avoid melting chip substrates.

2. Fermat's Theorem and the Converse Trap

Mathematical Foundation: Fermat's Theorem states that if $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c) = 0$ . However, students must avoid the classical Critical Number Converse Trap: $f'(c) = 0$ does not guarantee an extremum. For example, $f(x) = x^3$ has $f'(0) = 0$ , yet $(0, 0)$ is a flat inflection terrace—the function climbs continuously. To classify whether a critical number is a true local maximum, local minimum, or terrace, engineers construct a derivative sign chart and apply the First Derivative Test.

3. The Extreme Value Theorem & The Power of Preconditions

Mathematical Foundation: Before investing computational effort into finding an optimum, engineers must confirm that an optimal solution actually exists. The Extreme Value Theorem (EVT) guarantees that if a function is continuous on a closed interval $[a, b]$ , it is mathematically guaranteed to attain both an absolute maximum and an absolute minimum. If either precondition fails (e.g., an open boundary marker or a vertical asymptote inside the interval), the guarantee vanishes, and the system may grow unbounded.

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Skill Block 1

Real-Life Engineering Applications

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Electrical Engineering

Microchip Transistor Thermal Cycle Profiling (Electrical & Computer Engineering)

In high-power embedded electronics, autonomous robotics controllers, and server power supplies, gallium-nitride (GaN) high-electron-mobility transistors (HEMTs) switch high currents at megahertz frequencies. Rapid pulse-width modulation (PWM) switching induces dynamic thermal dissipation across the silicon-substrate junction.

An electronic systems engineer tests the thermal profile of a GaN power transistor during a scheduled 6-second operational duty cycle. Over the closed test interval $t \in [0, 6]$ seconds, the internal junction temperature $T$ (in degrees Celsius, $^\circ\text{C}$ ) is accurately modeled by the cubic polynomial:

T(t) = t^3 - 9t^2 + 15t + 100 \quad \text{for} \quad 0 \le t \le 6

To ensure hardware reliability and prevent catastrophic component failure, the hardware design team must audit the thermal operating limits against strict safety specifications:

The maximum junction temperature rating of the silicon package is $T_{\text{fail}} = 125^\circ\text{C}$ . Exceeding this thermal ceiling triggers thermal runaway, melting internal bond wires.
High-precision analog-to-digital signal calibration routines require minimal thermal noise interference, which occurs when the transistor is at its coolest operating point during the active cycle.

Locate all critical numbers of the junction temperature function $T(t)$ on the closed interval $[0, 6]$ , and classify each critical number using the First Derivative Test.
Determine the absolute maximum temperature and absolute minimum temperature reached during the duty cycle, identifying both the extreme values and their exact timing locations.
Evaluate the component safety margin ( $T_{\text{fail}} - T_{\max}$ ) and explain at what exact time instant firmware engineers should schedule precision sensor calibration to minimize thermal electronic noise.

💡 Show Worked Derivation

Problem Overview

We analyze the junction temperature $T(t) = t^3 - 9t^2 + 15t + 100$ ( $^\circ\text{C}$ ) of a power GaN transistor during a $6$ -second cycle ( $t \in [0, 6]$ ). We locate critical numbers, apply the First Derivative Test, determine absolute extrema using the Closed Interval Method, and audit hardware thermal safety and firmware calibration timing.

Step-by-Step Mathematical Solution

Step 1: Differentiate and Locate Critical Numbers

The governing formulation is the definition of Critical Numbers:

\frac{dT}{dt} = 0 \quad \text{or} \quad \frac{dT}{dt} \text{ does not exist}

(This concept is defined in Critical Numbers and Critical Points.)

Compute the first derivative using the Power Rule: $T'(t) = \frac{d}{dt}[t^3 - 9t^2 + 15t + 100] = 3t^2 - 18t + 15$
Set $T'(t) = 0$ to find stationary points: $3(t^2 - 6t + 5) = 0$ Factoring the quadratic: $3(t - 1)(t - 5) = 0 \implies t = 1\text{ s} \quad \text{and} \quad t = 5\text{ s}$
Check singular points: Since $T'(t)$ is a polynomial, it is defined for all $t$ .
Both $t = 1\text{ s}$ and $t = 5\text{ s}$ lie strictly within the interior interval $(0, 6)$ .

Step 2: Classify Critical Numbers via First Derivative Test

Construct a sign chart for $T'(t) = 3(t - 1)(t - 5)$ across the interval $[0, 6]$ :

Interval $(0, 1)$ : Test $t = 0.5 \implies T'(0.5) = 3(-0.5)(-4.5) = +6.75 > 0 \implies (+)$ . Temperature is increasing ( $\nearrow$ ).
Interval $(1, 5)$ : Test $t = 2 \implies T'(2) = 3(1)(-3) = -9 < 0 \implies (-)$ . Temperature is decreasing ( $\searrow$ ).
Interval $(5, 6)$ : Test $t = 5.5 \implies T'(5.5) = 3(4.5)(0.5) = +6.75 > 0 \implies (+)$ . Temperature is increasing ( $\nearrow$ ).

Classification:

At $t = 1\text{ s}$ : $T'(t)$ changes sign from positive to negative ( $+ \to -$ ). By the First Derivative Test, $t = 1\text{ s}$ is a local maximum.
At $t = 5\text{ s}$ : $T'(t)$ changes sign from negative to positive ( $- \to +$ ). By the First Derivative Test, $t = 5\text{ s}$ is a local minimum.

Step 3: Evaluate Candidate Values for Absolute Extrema

Using the Closed Interval Method Algorithm:

Left Boundary ( $t = 0\text{ s}$ ): $T(0) = (0)^3 - 9(0)^2 + 15(0) + 100 = 100^\circ\text{C}$
First Critical Point ( $t = 1\text{ s}$ ): $T(1) = (1)^3 - 9(1)^2 + 15(1) + 100 = 1 - 9 + 15 + 100 = 107^\circ\text{C}$
Second Critical Point ( $t = 5\text{ s}$ ): $T(5) = (5)^3 - 9(5)^2 + 15(5) + 100 = 125 - 225 + 75 + 100 = 75^\circ\text{C}$
Right Boundary ( $t = 6\text{ s}$ ): $T(6) = (6)^3 - 9(6)^2 + 15(6) + 100 = 216 - 324 + 90 + 100 = 82^\circ\text{C}$

Comparison:

Absolute Maximum Temperature Value: $107^\circ\text{C}$ at location $t = 1\text{ second}$ .
Absolute Minimum Temperature Value: $75^\circ\text{C}$ at location $t = 5\text{ seconds}$ .

Step 4: Engineering Analysis & Diagnostic Implications

Safety Margin Audit: The component thermal ceiling is $T_{\text{fail}} = 125^\circ\text{C}$ . The peak temperature is $T_{\max} = 107^\circ\text{C}$ . $\text{Safety Margin} = T_{\text{fail}} - T_{\max} = 125^\circ\text{C} - 107^\circ\text{C} = 18^\circ\text{C}$ Since the margin is positive ( $18^\circ\text{C} > 0$ ), the transistor operates safely below its thermal breakdown threshold throughout the duty cycle.
Firmware Calibration Scheduling: Johnson-Nyquist electronic thermal noise in semiconductor junctions is proportional to absolute temperature ( $v_n^2 = 4k_B T R \Delta f$ ). The global minimum temperature of $75^\circ\text{C}$ occurs at $t = 5\text{ seconds}$ . Firmware engineers should program the microcontroller to trigger high-precision ADC calibration sequences during the window surrounding $t = 5\text{ seconds}$ to ensure maximum signal-to-noise ratio (SNR).

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Critical Numbers, Fermat's Theorem, and Local Extrema Classification

General Overview
Skill Block 1 establishes the mathematical grammar of optimization. Students learn to rigorously distinguish between an extreme value ( $y$ -coordinate) and its location ( $x$ -coordinate), enforce the Endpoint Rule (endpoints can never be relative extrema), locate critical numbers where $f'(c) = 0$ or $f'(c)$ is undefined, and avoid mistaking vertical asymptotes for critical numbers.

Conceptual Framework

Extreme Values vs. Locations:
- The location is the input coordinate $x = c$ .
- The extreme value is the function output height $y = f(c)$ .
- The extreme point is the ordered pair $(c, f(c))$ on the Cartesian grid.
The Endpoint Rule:
- A relative (local) extremum requires an open neighborhood $(c - \delta, c + \delta)$ centered at $c$ that lies entirely within the domain of $f$ .
- Because interval boundaries have domain points on only one side, domain endpoints can qualify as absolute extrema, but they are strictly disqualified from being relative extrema.
The Domain Requirement for Critical Numbers:
- A number $c$ is a critical number if and only if $c \in \text{dom}(f)$ AND ( $f'(c) = 0$ or $f'(c) \text{ DNE}$ ).
- Points of discontinuity (such as $x = 0$ for $f(x) = \frac{1}{x}$ ) are not critical numbers because they are excluded from the domain.

Key Equations & Structural Rules

Critical Number Condition: $c \in \text{dom}(f) \quad \text{and} \quad [f'(c) = 0 \quad \text{or} \quad f'(c) \text{ DNE}]$ (see Critical Numbers and Critical Points)
Fermat's Theorem on Local Extrema: $\text{Local extremum at } c \implies f'(c) = 0 \quad (\text{if } f'(c) \text{ exists})$ (see Absolute and Relative Extrema)

Worked Example A: Classifying Extrema and Critical Numbers from Graphical Geometry

Given the graph of $f(x)$ below defined over the closed interval $[a, e]$ , identify the locations and values of all absolute and relative extrema.

12 - Relative Extrema Guided Notes Image 1.png

💡 Reveal Worked Solution

Name the Governing Rules: The Definitions of Absolute and Relative Extrema and the Endpoint Rule: (These definitions are formalized in Absolute and Relative Extrema and Critical Numbers and Critical Points.)
Identify Absolute Extrema (Global Highest and Lowest Heights):
- Absolute Maximum: Occurs at location $x = d$ . The curve reaches its greatest vertical height across the entire domain $[a, e]$ at the peak $(d, f(d))$ .
- Absolute Minimum: Occurs at location $x = a$ . The lowest point on the entire curve occurs at the left boundary endpoint $(a, f(a))$ .
Identify Relative (Local) Extrema (Interior Peaks and Valleys):
- Relative Maxima: Occur at locations $x = b$ and $x = d$ . In an open neighborhood around $x = b$ , $f(b) \ge f(x)$ , forming a local peak. Similarly, at $x = d$ , $f(d)$ is greater than all surrounding points.
- Relative Minimum: Occurs at location $x = c$ . In an open interval around $x = c$ , $f(c) \le f(x)$ , forming a local trough.
Enforce the Endpoint Rule:
- The boundaries $x = a$ and $x = e$ are endpoints. Because they do not possess a two-sided open neighborhood within $[a, e]$ , they cannot be classified as relative extrema. The point $x = a$ is strictly an absolute minimum, and $x = e$ is neither.

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) In the Conceptual Framework, state the "Endpoint Rule." If a continuous function has its highest point at a boundary endpoint $x = b$ , can $x = b$ be classified as a relative maximum?
(b) Deconstruct Worked Example A: Why is $x = d$ classified as both a relative maximum and an absolute maximum, whereas $x = b$ is classified only as a relative maximum?
(c) Review the linked Engineering Application (Microchip Transistor Thermal Cycle Profiling (Electrical & Computer Engineering)). Why must electronic systems engineers calculate both the maximum value and the minimum value of junction temperature during a duty cycle?

Practice Space Solve this problem on paper or tablet.

Review the Endpoint Rule in the Conceptual Framework, compare heights at $x = b$ and $x = d$ in Worked Example A, and read the Problem Overview and Step 4 of Microchip Transistor Thermal Cycle Profiling (Electrical & Computer Engineering).

Answer: * (a) Relative extrema require a two-sided open neighborhood within the domain; endpoints have points on only one side and therefore can never be relative extrema.

(b) $x = d$ is an interior peak with a two-sided neighborhood (local max) and also achieves the highest function value on the entire interval (absolute max); $x = b$ is a local peak but is exceeded by $f(d)$ .
(c) The maximum temperature ( $107^\circ\text{C}$ ) verifies safety against the thermal melting ceiling ( $125^\circ\text{C}$ ); the minimum temperature ( $75^\circ\text{C}$ ) identifies the exact low-noise timing window for sensor calibration.

Step-by-Step Execution

(a) The Endpoint Rule states that relative (local) extrema can never occur at the endpoints of an interval because the definition of a relative extremum requires an open interval $(c - \delta, c + \delta)$ centered at $c$ lying entirely inside the domain. At an endpoint, function values exist on only one side. Therefore, while $x = b$ can be the absolute maximum of the function on $[a, b]$ , it cannot be classified as a relative maximum.
(b) In Worked Example A, location $x = d$ lies in the interior of the interval $[a, e]$ and is a turning point where $f(d)$ is larger than all function values in an immediate two-sided open neighborhood, qualifying it as a relative maximum. Simultaneously, $f(d)$ is the greatest function value across the entire interval, making it the absolute maximum. Location $x = b$ is also an interior turning point higher than its immediate neighbors (so it is a relative maximum), but because $f(b) < f(d)$ , it does not attain the highest global value on the domain.
(c) In GaN power transistor design, calculating the absolute maximum junction temperature ( $107^\circ\text{C}$ at $t = 1\text{ s}$ ) confirms that the device maintains a safe $18^\circ\text{C}$ margin below its catastrophic melting ceiling of $125^\circ\text{C}$ . Calculating the absolute minimum temperature ( $75^\circ\text{C}$ at $t = 5\text{ s}$ ) identifies the point of lowest thermal electronic noise ( $v_n \propto \sqrt{T}$ ), enabling firmware engineers to synchronize high-precision sensor calibration readings to this optimal temporal window.

Practice Problem 2

Consider the graph below defined over $[-3, 6]$ :

12 - Absolute and Relative Extrema Image 1.png

Identify all relative and absolute extrema (stating both values and locations), or explain why they do not exist:

(a) Relative Maxima
(b) Absolute Maximum
(c) Relative Minima
(d) Absolute Minimum

Practice Space Solve this problem on paper or tablet.

Review the definitions of Relative Extrema, Absolute Extrema, and the Endpoint Rule. Pay close attention to open circles versus solid dots on boundaries and branch terminations.

Answer: * (a) Relative Maxima: Value $3$ at location $x = 0$ .

(b) Absolute Maximum: Value $5$ at locations $x = 2$ and $x = 6$ .
(c) Relative Minima: Value $-1$ at location $x = -2$ , and Value $1$ at location $x = 4$ .
(d) Absolute Minimum: Does not exist (DNE).

Step-by-Step Execution

Name the Governing Rules: Definition of Relative and Absolute Extrema: (These definitions are formalized in Absolute and Relative Extrema.)
Evaluate Relative Maxima:
- At $x = 0$ , the curve reaches an interior peak with value $f(0) = 3$ . This is an interior turning point with a complete two-sided neighborhood, so the relative maximum value is $3$ at $x = 0$ .
- The endpoints $x = -3$ and $x = 6$ are disqualified by the Endpoint Rule. The point $x = 2$ has domain values only for $x \ge 2$ because the left branch terminates at $x \approx 1.25$ , so it has no two-sided neighborhood.
Evaluate Absolute Maximum:
- The greatest function height reached on the graph is $y = 5$ .
- This value is attained at two locations: $x = 2$ and $x = 6$ . Both coordinates have solid dots, confirming that the value $5$ is achieved. Absolute maximum value is $5$ at $x = 2$ and $x = 6$ .
Evaluate Relative Minima:
- At $x = -2$ , the curve reaches an interior valley with value $f(-2) = -1$ .
- At $x = 4$ , the curve reaches an interior valley with value $f(4) = 1$ .
- Both locations possess two-sided open neighborhoods where the point is lower than its immediate neighbors. Relative minima values are $-1$ at $x = -2$ and $1$ at $x = 4$ .
Evaluate Absolute Minimum:
- The graph approaches a low vertical height of $-4$ as $x \to 1.25$ on the left branch, but terminates in an open circle at $(1.25, -4)$ .
- Because this boundary point is excluded, the function values get arbitrarily close to $-4$ but never reach it. Since the function dips below all other valleys but never attains a minimum value, the absolute minimum does not exist (DNE).

Practice Problem 3

Find all critical numbers of the rational function on its domain:

f(x) = \frac{x^2}{x - 2}

Practice Space Solve this problem on paper or tablet.

State the Quotient Rule formula. Set $f'(x) = 0$ to find stationary points and check where $f'(x)$ is undefined. Verify each candidate against the domain of the original function.

Answer:

\text{Critical numbers: } x = 0 \quad \text{and} \quad x = 4 \quad (x = 2 \text{ is excluded from domain})

Step-by-Step Execution

Name the Governing Formula: Quotient Rule and Definition of Critical Numbers: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$ $c \in \text{dom}(f) \quad \text{such that} \quad f'(c) = 0 \quad \text{or} \quad f'(c) \text{ DNE}$ (These definitions are formalized in Differentiation Rules - Basics, Powers, Products and Quotients and Critical Numbers and Critical Points.)
Determine the Domain: The rational function $f(x)$ is defined for all real numbers except where the denominator is zero: $x - 2 \neq 0 \implies x \neq 2$ $\text{dom}(f) = (-\infty, 2) \cup (2, \infty)$
Compute the First Derivative: Let $u(x) = x^2 \implies u'(x) = 2x$ , and $v(x) = x - 2 \implies v'(x) = 1$ : $f'(x) = \frac{2x(x - 2) - x^2(1)}{(x - 2)^2} = \frac{2x^2 - 4x - x^2}{(x - 2)^2} = \frac{x^2 - 4x}{(x - 2)^2} = \frac{x(x - 4)}{(x - 2)^2}$
Locate Stationary Points ( $f'(x) = 0$ ): Set the numerator equal to zero: $x(x - 4) = 0 \implies x = 0 \quad \text{or} \quad x = 4$ Both $x = 0$ and $x = 4$ lie in the domain of $f(x)$ .
Check Singular Points ( $f'(x) \text{ DNE}$ ): The derivative is undefined when its denominator is zero: $(x - 2)^2 = 0 \implies x = 2$ However, $x = 2$ is NOT in the domain of $f(x)$ ( $f(2)$ is undefined; $x = 2$ is a vertical asymptote). Therefore, $x = 2$ cannot be a critical number.
Conclusion: The critical numbers are strictly $x = 0$ and $x = 4$ .

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📅 Regroup 1

Fermat's Theorem Converse Misconception: Emphasize that finding $f'(c) = 0$ does not guarantee a local extremum. Show the cubic counterexample $f(x) = x^3$ at $x = 0$ .
The Domain Precondition: Reinforce that vertical asymptotes can never be critical numbers because critical numbers must belong to the domain of the function.
Precision in Reporting: Review student answers to ensure they state $x$ -values for locations and $y$ -values for extreme values.

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