MATH-181 (Calculus I) Curriculum

Session 13 - Critical Points and Extrema

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Locate and classify critical numbers of polynomial, rational, and radical functions where $f'(c) = 0$ or $f'(c)$ is undefined ( $c \in \text{dom}(f)$ ), distinguishing between extreme values ( $y$ ) and locations ( $x$ ), and applying Fermat's Theorem while recognizing that $f'(c) = 0$ is a necessary but not sufficient condition for local extrema.
Construct 1D derivative sign charts to analyze intervals of increase and decrease, and execute the First Derivative Test to classify local extrema and determine absolute extrema on open or unbounded intervals via the Single Critical Point Rule.
Diagnose the hypotheses of the Extreme Value Theorem (EVT) (continuity on a closed interval $[a, b]$ ) and execute the 5-step Closed Interval Method Algorithm to locate absolute maxima and absolute minima across algebraic, logarithmic, and trigonometric functions.

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.

Skill Block 1

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

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?

Solution Key:

(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.

Show your steps and calculations here.

Practice Problem 2

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

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

12 - Absolute and Relative Extrema Image 1.png

Solution Key:

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).

Show your steps and calculations here.

Practice Problem 3

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

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

Solution Key:

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 $x = 4$ .

Show your steps and calculations here.

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.

Skill Block 2

Worked Example A: First Derivative Test and the Single Critical Point Rule on an Open Domain

Find and classify all local and absolute extrema of the function:

f(x) = x - \ln(x) \quad \text{on its natural domain } (0, \infty)

Worked Solution:

Name the Governing Rule: The First Derivative Test and Single Critical Point Rule: (These rules are defined in First Derivative Test.)
Compute the First Derivative: Differentiating term by term: $f'(x) = \frac{d}{dx}[x] - \frac{d}{dx}[\ln x] = 1 - \frac{1}{x} = \frac{x - 1}{x}$
Locate Critical Numbers:
- Stationary point ( $f'(x) = 0$ ): $\frac{x - 1}{x} = 0 \implies x - 1 = 0 \implies x = 1$
- Singular point ( $f'(x) \text{ DNE}$ ): $x = 0$ , but $0 \notin (0, \infty)$ .
- Thus, $x = 1$ is the sole interior critical number.
Construct the 1D Sign Chart: Partition $(0, \infty)$ at $x = 1$ into two test intervals:
- Interval $(0, 1)$ : Choose test value $x = 0.5$ : $f'(0.5) = 1 - \frac{1}{0.5} = 1 - 2 = -1 < 0 \implies (-)$ $f(x)$ is decreasing on $(0, 1)$ .
- Interval $(1, \infty)$ : Choose test value $x = 2$ : $f'(2) = 1 - \frac{1}{2} = 0.5 > 0 \implies (+)$ $f(x)$ is increasing on $(1, \infty)$ .
Classify Extrema & Apply Single Critical Point Rule:
- Because $f'(x)$ changes sign from negative to positive ( $- \to +$ ) at $x = 1$ , $f(1)$ is a local minimum.
- By the Single Critical Point Rule, because $x = 1$ is the only critical number in the entire open interval $(0, \infty)$ , this local minimum is strictly the absolute minimum: $f(1) = 1 - \ln(1) = 1 - 0 = 1$
- Absolute Minimum Value: $1$ at location $x = 1$ .
- Absolute Maximum: Does not exist (as $x \to 0^+$ , $-\ln x \to \infty$ ; as $x \to \infty$ , $x \to \infty$ ).

Practice Problem 4: Concept & Reading Check

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

(a) In the Conceptual Framework, explain why test values must be substituted strictly into the derivative $f'(x)$ rather than the original function $f(x)$ during sign analysis.
(b) Deconstruct Worked Example A: Why does the presence of only a single critical number on $(0, \infty)$ allow us to conclude that $f(1) = 1$ is an absolute minimum without needing to test endpoints?
(c) Review the linked Engineering Application (Chemical Bioreactor Dissolved Oxygen Depletion (Chemical & Environmental Engineering)). At what time does dissolved oxygen peak, and what mathematical theorem justifies that this peak is global?

Solution Key:

(a) Sign analysis determines the directional trajectory of a curve (whether it is climbing or falling). By the Monotonicity Theorem, the algebraic sign of the first derivative $f'(x)$ dictates this slope ( $f' > 0$ means increasing; $f' < 0$ means decreasing). Substituting a test value into the original function $f(x)$ produces only the vertical height of the curve at that point, which provides zero information regarding whether the curve is sloping upward or downward.
(b) In Worked Example A, the domain is the open interval $(0, \infty)$ . Because $f(x)$ is continuous and $x = 1$ is the only critical number, the derivative $f'(x)$ cannot change sign anywhere else. Since $f'(x) < 0$ for all $x \in (0, 1)$ and $f'(x) > 0$ for all $x \in (1, \infty)$ , the curve descends monotonically to $x = 1$ and ascends monotonically forever after. It is geometrically impossible for the function to turn back down to achieve a lower value, guaranteeing that the local minimum at $x = 1$ is the global (absolute) minimum.
(c) In the wastewater bioreactor model, dissolved oxygen reaches its maximum concentration of $C(2) = 2.0\text{ mg/L}$ at location $t = 2\text{ hours}$ . Because $t = 2$ is the unique interior critical number on the operational domain $[0, \infty)$ and $C'(t)$ transitions from positive to negative, the Single Critical Point Rule rigorously proves that $2.0\text{ mg/L}$ is the absolute maximum concentration attainable before supplemental blowers must restart.

Show your steps and calculations here.

Practice Problem 5

Find and classify all local extrema and determine any absolute extrema for the polynomial:

g(x) = 5x^6 - 6x^5 + 3 \quad \text{on } (-\infty, \infty)

Solution Key:

Name the Governing Formula: Power Rule and First Derivative Test: $\frac{d}{dx}[x^n] = n x^{n-1}$ (These rules are defined in Differentiation Rules - Basics, Powers, Products and Quotients and First Derivative Test.)
Compute the Derivative and Factor: $g'(x) = 30x^5 - 30x^4 = 30x^4(x - 1)$
Locate Critical Numbers: Setting $g'(x) = 0$ : $30x^4(x - 1) = 0 \implies x = 0 \quad \text{and} \quad x = 1$
Construct 1D Sign Chart: The critical numbers partition $(-\infty, \infty)$ into $(-\infty, 0)$ , $(0, 1)$ , and $(1, \infty)$ . Note that $30x^4 > 0$ for all $x \neq 0$ :
- Interval $(-\infty, 0)$ : Test $x = -1 \implies g'(-1) = 30(-1)^4(-2) = -60 < 0 \implies (-)$ . Decreasing.
- Interval $(0, 1)$ : Test $x = 0.5 \implies g'(0.5) = 30(0.5)^4(-0.5) < 0 \implies (-)$ . Decreasing.
- Interval $(1, \infty)$ : Test $x = 2 \implies g'(2) = 30(2)^4(1) = 480 > 0 \implies (+)$ . Increasing.
Classify Extrema:
- At $x = 0$ : $g'(x)$ is negative on both sides. The curve flattens horizontally but continues falling. Thus, $(0, 3)$ is a terrace point (inflection point), neither a maximum nor a minimum.
- At $x = 1$ : $g'(x)$ changes sign from negative to positive ( $- \to +$ ). Thus, $g(1)$ is a local minimum: $g(1) = 5(1)^6 - 6(1)^5 + 3 = 5 - 6 + 3 = 2$
Determine Absolute Extrema via End Behavior: Because $g(x)$ is an even-degree polynomial with a positive leading coefficient ( $5x^6$ ): $\lim_{x \to \pm\infty} g(x) = \infty$ Because the function diverges to $+\infty$ on both boundaries, no absolute maximum exists. The local minimum of $2$ at $x = 1$ is the absolute minimum.

Show your steps and calculations here.

Practice Problem 6

Find and classify all critical points and absolute extrema of the function:

h(x) = 7 - (5 - 2x)^{2/3} \quad \text{on } (-\infty, \infty)

Solution Key:

Name the Governing Formula: Chain Rule and Definition of Singular Critical Numbers: $\frac{d}{dx}[u^{2/3}] = \frac{2}{3}u^{-1/3} \cdot u'$ (These rules are defined in Chain Rule and Critical Numbers and Critical Points.)
Compute the First Derivative: $h'(x) = -\frac{2}{3}(5 - 2x)^{-1/3} \cdot \frac{d}{dx}[5 - 2x] = -\frac{2}{3(5 - 2x)^{1/3}} \cdot (-2) = \frac{4}{3(5 - 2x)^{1/3}}$
Locate Critical Numbers:
- Stationary Points ( $h'(x) = 0$ ): The numerator is the constant $4 \neq 0$ . Therefore, $h'(x) \neq 0$ everywhere.
- Singular Points ( $h'(x) \text{ DNE}$ ): The derivative is undefined when the denominator is zero: $3(5 - 2x)^{1/3} = 0 \implies 5 - 2x = 0 \implies x = \frac{5}{2}$
- Since $h(5/2) = 7 - (0)^{2/3} = 7$ , $x = \frac{5}{2}$ belongs to the domain and is a valid critical number (a sharp cusp).
Construct 1D Sign Chart:
- Interval $(-\infty, 5/2)$ : Test $x = 0 \implies h'(0) = \frac{4}{3(5)^{1/3}} > 0 \implies (+)$ . Increasing ( $\nearrow$ ).
- Interval $(5/2, \infty)$ : Test $x = 3 \implies h'(3) = \frac{4}{3(5 - 6)^{1/3}} = \frac{4}{3(-1)} = -\frac{4}{3} < 0 \implies (-)$ . Decreasing ( $\searrow$ ).
Classify Extrema & End Behavior:
- Since $h'(x)$ transitions from $(+)$ to $(-)$ at $x = \frac{5}{2}$ , $h(5/2) = 7$ is a local maximum.
- As $x \to \pm\infty$ , $(5 - 2x)^{2/3} \to +\infty$ because the power of $2/3$ squares the quantity before taking the cube root. Thus: $\lim_{x \to \pm\infty} h(x) = 7 - \infty = -\infty$
- Since the curve descends infinitely on both ends, no absolute minimum exists.
- Absolute Maximum Value: $7$ at location $x = \frac{5}{2}$ .

Show your steps and calculations here.

Regroup 2

Testing Strictly in $f'(x)$ : Reiterate that when checking signs on a number line, test values must be substituted strictly into the derivative, not the original function.
Handling Singular Critical Points: Emphasize that sharp cusps ( $f'(c) \text{ DNE}$ ) can be absolute extrema just like smooth stationary points ( $f'(c) = 0$ ).
Single Critical Point Verification: Confirm students understand why the Single Critical Point Rule only works when there is exactly one critical point on the interval.

Skill Block 3

Worked Example A: Closed Interval Method on a Cubic Polynomial

Find the absolute maximum and absolute minimum values of the function:

f(x) = x^3 - 3x^2 + 1 \quad \text{on the closed interval } [-1, 4]

Worked Solution:

Name the Governing Algorithm: The Closed Interval Method Algorithm: (This procedure is defined in Closed Interval Method and Extreme Value Theorem.)
Step 1: Check Preconditions: $f(x)$ is a cubic polynomial, which is continuous for all real numbers $\mathbb{R}$ . The domain $[-1, 4]$ is a closed, bounded interval. The Extreme Value Theorem is satisfied; absolute extrema are guaranteed.
Step 2: Differentiate: $f'(x) = 3x^2 - 6x = 3x(x - 2)$
Step 3: Locate and Filter Interior Critical Numbers: Set $f'(x) = 0$ : $3x(x - 2) = 0 \implies x = 0 \quad \text{and} \quad x = 2$ Filter Check: Both $x = 0$ and $x = 2$ lie strictly within the interior open interval $(-1, 4)$ . Both are retained.
Step 4: Evaluate the Original Function $f(x)$ at All Candidates:
- Left Endpoint ( $x = -1$ ): $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3$
- First Critical Point ( $x = 0$ ): $f(0) = (0)^3 - 3(0)^2 + 1 = 1$
- Second Critical Point ( $x = 2$ ): $f(2) = (2)^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$
- Right Endpoint ( $x = 4$ ): $f(4) = (4)^3 - 3(4)^2 + 1 = 64 - 48 + 1 = 17$
Step 5: Compare and Conclude: Comparing candidate values $\{-3, 1, -3, 17\}$ :
- Absolute Maximum Value: $17$ at location $x = 4$ .
- Absolute Minimum Value: $-3$ at locations $x = -1$ and $x = 2$ .

Practice Problem 7: 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 two required hypotheses of the Extreme Value Theorem. If a function is continuous on the interval $(1, 5]$ , is it guaranteed to have an absolute minimum?
(b) Deconstruct Worked Example A: Why did we evaluate $f(-1)$ and $f(4)$ even though $f'(x) \neq 0$ at those points? What error occurs if an engineer evaluates $f'(0)$ and $f'(2)$ instead of $f(0)$ and $f(2)$ in Step 4?
(c) Review the linked Engineering Application (Structural Beam Bending Moment Optimization (Civil & Construction Engineering)). At what location does the maximum bending moment occur, and how does this calculus result prevent construction waste?

Solution Key:

(a) The two hypotheses of the Extreme Value Theorem are: (1) The function must be continuous on the interval, and (2) The interval must be closed and bounded ( $[a, b]$ ). If a function is defined on the half-open interval $(1, 5]$ , it violates the closed interval hypothesis. As the function approaches $x = 1$ , it may drop toward a hole or asymptote without ever reaching a lowest included point. Therefore, an absolute minimum is not guaranteed to exist.
(b) In Worked Example A, endpoints $x = -1$ and $x = 4$ must be evaluated because the boundaries of a physical domain represent hard operational limits where the function may attain its global highest or lowest point without the derivative flattening to zero. If an engineer mistakenly substitutes critical numbers into the derivative $f'(x)$ instead of the original function $f(x)$ , they obtain $f'(0) = 0$ and $f'(2) = 0$ . This reports zero slope rather than the actual physical values ( $1$ and $-3$ ), invalidating the entire comparison.
(c) In structural beam design, the internal bending moment peaks at $M(4) = 66\text{ kN}\cdot\text{m}$ at location $x = 4\text{ meters}$ . Because tensile flexural cracking initiates where bending moment peaks, civil engineers place dense steel rebar reinforcement specifically at the 4-meter mark. At the right support wall ( $x = 6\text{ meters}$ ), the moment drops to $14\text{ kN}\cdot\text{m}$ . Tapering rebar reinforcement to match this moment distribution prevents the waste of tons of structural steel that would occur if maximum reinforcement were uniformly installed across the entire span.

Show your steps and calculations here.

Practice Problem 8

Determine whether the Extreme Value Theorem applies to the following functions on the given intervals. If it does not apply, identify which specific hypothesis failed:

(a) $f(x) = \frac{x + 2}{x - 3} \quad \text{on } [1, 5]$
(b) $g(x) = 4x^3 - 12x \quad \text{on } (0, 3]$

Solution Key:

Name the Governing Theorem: Extreme Value Theorem Precondition Criteria: (This theorem is defined in Extreme Value Theorem.)
Analysis for Part (a):
- Interval Check: $[1, 5]$ uses square brackets, so it is a closed interval.
- Continuity Check: $f(x) = \frac{x + 2}{x - 3}$ has a denominator that equals zero when $x - 3 = 0 \implies x = 3$ . Because $x = 3$ is in the interior of the interval $[1, 5]$ , $f(x)$ has an infinite discontinuity (vertical asymptote) inside the domain.
- Conclusion for (a): The EVT does not apply because $f(x)$ fails the hypothesis of continuity on $[1, 5]$ .
Analysis for Part (b):
- Continuity Check: $g(x) = 4x^3 - 12x$ is a polynomial, which is continuous everywhere on $\mathbb{R}$ .
- Interval Check: The domain is $(0, 3]$ . The left boundary at $x = 0$ is indicated by a parenthesis, which excludes $x = 0$ .
- Conclusion for (b): The EVT does not apply because the interval is not closed.

Show your steps and calculations here.

Practice Problem 9

Find the absolute maximum and absolute minimum values and their locations for:

f(x) = \ln(x^2 + 3x + 4) \quad \text{on the closed interval } [-2, 2]

Solution Key:

Name the Governing Algorithm: Closed Interval Method with Chain Rule: $\frac{d}{dx}[\ln(u(x))] = \frac{u'(x)}{u(x)}$ (This procedure is defined in Closed Interval Method and Chain Rule.)
Step 1: Check Preconditions: The argument of the logarithm is $x^2 + 3x + 4$ . Its discriminant is $\Delta = b^2 - 4ac = 9 - 16 = -7 < 0$ . Because the discriminant is negative and the leading coefficient is positive, $x^2 + 3x + 4 > 0$ for all real $x$ . Thus, $f(x)$ is continuous everywhere on the closed interval $[-2, 2]$ .
Step 2: Differentiate: $f'(x) = \frac{2x + 3}{x^2 + 3x + 4}$
Step 3: Locate and Filter Interior Critical Numbers: Set $f'(x) = 0$ : $2x + 3 = 0 \implies x = -\frac{3}{2} = -1.5$ Since $-1.5 \in (-2, 2)$ , it is a valid interior critical number. (The denominator is never zero, so there are no singular points).
Step 4: Evaluate $f(x)$ at Candidates:
- Left Endpoint ( $x = -2$ ): $f(-2) = \ln((-2)^2 + 3(-2) + 4) = \ln(4 - 6 + 4) = \ln(2) \approx 0.693$
- Interior Critical Point ( $x = -3/2$ ): $f\left(-\frac{3}{2}\right) = \ln\left(\left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) + 4\right) = \ln\left(\frac{9}{4} - \frac{9}{2} + 4\right) = \ln\left(\frac{7}{4}\right) \approx 0.560$
- Right Endpoint ( $x = 2$ ): $f(2) = \ln((2)^2 + 3(2) + 4) = \ln(4 + 6 + 4) = \ln(14) \approx 2.639$
Step 5: Compare Results:
- Absolute Maximum Value: $\ln(14) \approx 2.639$ at location $x = 2$ .
- Absolute Minimum Value: $\ln\left(\frac{7}{4}\right) \approx 0.560$ at location $x = -\frac{3}{2}$ .

Show your steps and calculations here.

Practice Problem 10

Find the absolute maximum and absolute minimum values and their locations for:

f(x) = \cos\left(x + \frac{\pi}{2}\right) \quad \text{on the closed interval } \left[0, \frac{3\pi}{2}\right]

Solution Key:

Name the Governing Algorithm: Closed Interval Method for Trigonometric Functions: (This procedure is defined in Closed Interval Method.)
Step 1: Simplify and Check Preconditions: Using the co-function identity: $\cos\left(x + \frac{\pi}{2}\right) = -\sin(x)$ The sine function is continuous everywhere on the closed interval $[0, \frac{3\pi}{2}]$ .
Step 2 & 3: Differentiate and Find Critical Numbers: $f'(x) = -\cos(x)$ Set $f'(x) = 0$ on the interior interval $\left(0, \frac{3\pi}{2}\right)$ : $-\cos(x) = 0 \implies x = \frac{\pi}{2}$ $x = \frac{\pi}{2}$ is the only interior critical number in $\left(0, \frac{3\pi}{2}\right)$ .
Step 4: Evaluate $f(x)$ at Candidates:
- Left Endpoint ( $x = 0$ ): $f(0) = -\sin(0) = 0$
- Interior Critical Point ( $x = \frac{\pi}{2}$ ): $f\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$
- Right Endpoint ( $x = \frac{3\pi}{2}$ ): $f\left(\frac{3\pi}{2}\right) = -\sin\left(\frac{3\pi}{2}\right) = -(-1) = 1$
Step 5: Compare Results:
- Absolute Maximum Value: $1$ at location $x = \frac{3\pi}{2}$ .
- Absolute Minimum Value: $-1$ at location $x = \frac{\pi}{2}$ .

Show your steps and calculations here.

Practice Problem 11

Find the absolute maximum and absolute minimum values and their locations for:

g(x) = 2x - 3x^{2/3} \quad \text{on the closed interval } [-1, 8]

Solution Key:

Name the Governing Algorithm: Closed Interval Method with Singular Points: (This procedure is defined in Closed Interval Method and Critical Numbers and Critical Points.)
Step 1: Check Preconditions: $g(x) = 2x - 3(\sqrt[3]{x})^2$ is continuous everywhere, including on $[-1, 8]$ .
Step 2: Differentiate: $g'(x) = 2 - 3\left(\frac{2}{3}x^{-1/3}\right) = 2 - \frac{2}{x^{1/3}} = \frac{2x^{1/3} - 2}{x^{1/3}}$
Step 3: Locate and Filter Interior Critical Numbers:
- Stationary Points ( $g'(x) = 0$ ): Numerator equals zero: $2x^{1/3} - 2 = 0 \implies x^{1/3} = 1 \implies x = 1$
- Singular Points ( $g'(x) \text{ DNE}$ ): Denominator equals zero: $x^{1/3} = 0 \implies x = 0$
- Filter: Both $x = 0$ and $x = 1$ lie strictly inside the interior interval $(-1, 8)$ .
Step 4: Evaluate $g(x)$ at All Candidates:
- Left Endpoint ( $x = -1$ ): $g(-1) = 2(-1) - 3(-1)^{2/3} = -2 - 3(1) = -5$
- Singular Critical Point ( $x = 0$ ): $g(0) = 2(0) - 3(0)^{2/3} = 0$
- Stationary Critical Point ( $x = 1$ ): $g(1) = 2(1) - 3(1)^{2/3} = 2 - 3 = -1$
- Right Endpoint ( $x = 8$ ): $g(8) = 2(8) - 3(8)^{2/3} = 16 - 3(4) = 16 - 12 = 4$
Step 5: Compare Results: Comparing candidate heights $\{-5, 0, -1, 4\}$ :
- Absolute Maximum Value: $4$ at location $x = 8$ .
- Absolute Minimum Value: $-5$ at location $x = -1$ .

Show your steps and calculations here.

Regroup 3

Evaluating in $f(x)$ vs $f'(x)$ : Ensure every student understands why candidate heights are computed strictly in the original function $f(x)$ .
Singular Points on Closed Intervals: Remind students to always check where the derivative denominator equals zero (singular points like $x = 0$ in Problem 11).
The Role of Endpoints: Emphasize that endpoints are equal-status candidates in global optimization and must never be skipped.

Synthesis Wrap-up

Connecting the Toolkit: Today we transitioned from understanding local behavior (where slopes flatten or peak via Fermat's Theorem and the First Derivative Test) to global optimization on closed intervals via the Extreme Value Theorem.
Looking Ahead to Sessions 14 & 15: In upcoming sessions, we take this exact mathematical machinery and apply it to real-world geometric and kinematic design: finding the dimensions that maximize container volume or minimize pipeline construction cost (constrained optimization).
Homework Preparation: Complete the Knewton Alta assignment Critical Points and Extrema. Ensure you check your domain restrictions and test signs carefully!