Mathematics / Calculus I Single-Variable Differential Calculus 100% Free Open Access
Chapter 8 β€’ Theory & Derivations

Differential Curve Sketching, Extreme Values & Optimization

Comprehensive applications of derivatives: Fermat's interior extremum theorem, critical numbers, First and Second Derivative Tests, concavity and inflection points, universal 7-step analytical curve sketching algorithm, applied geometric and physical optimization problems, and the related rates framework.

Β§8.1 Critical Numbers, Fermat's Theorem on Stationary Points & Extrema

1. Local Extrema & Critical Numbers

$$\mathbf{\text{Definition (Critical Number): A number } c \in \operatorname{Dom}(f) \text{ is a critical number of } f \text{ if either:}}$$ $$\mathbf{f'(c) = 0 \quad \text{(stationary point)} \quad \text{or} \quad f'(c) \text{ is undefined (sharp corner, cusp, vertical tangent).}}$$

2. Fermat's Interior Extremum Theorem

$$\mathbf{\text{Theorem (Fermat): If } f \text{ has a local extremum (maximum or minimum) at an interior point } c, \text{ and } f'(c) \text{ exists, then } f'(c) = 0.}$$

Proof for Local Maximum: Suppose $f$ has a local maximum at $c$. Then there exists $\delta > 0$ such that $f(x) \le f(c)$ for all $x \in (c - \delta, c + \delta)$.

  • For $h \in (0, \delta)$ (approaching from right): $f(c + h) - f(c) \le 0 \implies \frac{f(c + h) - f(c)}{h} \le 0 \implies f'_+(c) \le 0$.
  • For $h \in (-\delta, 0)$ (approaching from left): $f(c + h) - f(c) \le 0 \implies \frac{f(c + h) - f(c)}{h} \ge 0 \implies f'_-(c) \ge 0$.

Since $f$ is differentiable at $c$, $f'(c) = f'_+(c) = f'_-(c)$. The only real number satisfying both $f'(c) \le 0$ and $f'(c) \ge 0$ is $f'(c) = 0 \quad \blacksquare$

3. The Closed Interval Method for Absolute Extrema

To find the absolute maximum and minimum of a continuous function $f$ on a closed bounded interval $[a, b]$:

  1. Find all critical numbers $c_i \in (a, b)$.
  2. Evaluate $f(c_i)$ at every critical number.
  3. Evaluate $f(a)$ and $f(b)$ at the endpoints.
  4. The greatest evaluated value is the absolute maximum; the least is the absolute minimum.

Β§8.2 First & Second Derivative Tests, Concavity & Inflection Points

1. The First Derivative Test for Local Extrema

Let $c$ be a critical number of a continuous function $f$:

  • If $f'(x)$ changes from positive to negative as $x$ increases through $c$, then $f(c)$ is a local maximum.
  • If $f'(x)$ changes from negative to positive as $x$ increases through $c$, then $f(c)$ is a local minimum.
  • If $f'(x)$ does not change sign across $c$ (e.g. $+ \to +$ or $- \to -$), then $f(c)$ is neither a local maximum nor a local minimum (e.g. $f(x) = x^3$ at $c = 0$).

2. Concavity & Inflection Points

$$\begin{aligned} \mathbf{\text{Concave Upward: }} & f''(x) > 0 \quad \forall x \in I \iff \text{The tangent lines lie strictly below the curve} \ \mathbf{\text{Concave Downward: }} & f''(x) < 0 \quad \forall x \in I \iff \text{The tangent lines lie strictly above the curve} \end{aligned}$$

Point of Inflection: A point $(c, f(c))$ on the curve where the function is continuous and the concavity changes (from upward to downward, or vice versa). A necessary condition for an inflection point is $f''(c) = 0$ or $f''(c)$ undefined.

3. The Second Derivative Test for Local Extrema

Suppose $f''(x)$ is continuous near a stationary point $c$ with $f'(c) = 0$:

  • If $f''(c) > 0$, then the curve is concave upward at $c$, so $f(c)$ is a local minimum.
  • If $f''(c) < 0$, then the curve is concave downward at $c$, so $f(c)$ is a local maximum.
  • If $f''(c) = 0$, the test is inconclusive (e.g. $x^4$ has a minimum, $-x^4$ has a maximum, $x^3$ has an inflection point); one must revert to the First Derivative Test.

Β§8.3 Systematic 7-Step Protocol for Mathematical Curve Sketching

The Universal 7-Step Curve Sketching Protocol

To sketch the graph of an arbitrary function $y = f(x)$ with analytical precision without guesswork:

  1. Step 1: Domain & Symmetries β€” Identify $\operatorname{Dom}(f)$. Test for even symmetry $f(-x) = f(x)$, odd symmetry $f(-x) = -f(x)$, or periodicity $f(x+T) = f(x)$.
  2. Step 2: Intercepts β€” $y$-intercept at $(0, f(0))$. $x$-intercepts by solving $f(x) = 0$.
  3. Step 3: Asymptotes β€” Vertical asymptotes where $Q(x) = 0$ with $\lim |f(x)| = \infty$. Horizontal asymptotes $\lim_{x \to \pm\infty} f(x) = L$. Slant asymptotes $y = mx + b$ if $\lim [f(x) - (mx+b)] = 0$.
  4. Step 4: First Derivative $f'(x)$ β€” Critical numbers ($f'(x) = 0$ or undefined). Sign chart for $f'(x)$ to determine intervals of increase and decrease.
  5. Step 5: Local Extrema β€” Apply First or Second Derivative Test to classify each critical point.
  6. Step 6: Second Derivative $f''(x)$ β€” Determine $f''(x) = 0$ or undefined. Sign chart for $f''(x)$ to determine intervals of concavity upward/downward and exact inflection points.
  7. Step 7: Global Assembly & Plot β€” Plot intercepts, asymptotes, extrema, and inflection points; sketch the smooth curve following the concavity and monotonicity signatures.

Β§8.4 Applied Real-World Optimization & Related Rates Framework

1. Applied Mathematical Optimization Protocol

  1. Variable Identification & Sketch: Assign variables to all quantities. Draw a diagram.
  2. Objective Function: Write an explicit formula for the quantity $Q$ to be maximized or minimized (e.g. volume, area, cost, material).
  3. Constraint Equations: Relate secondary variables via geometric or physical constraints (e.g. surface area budget, perimeter, Pythagorean theorem) to express $Q = f(x)$ purely in terms of a single independent variable.
  4. Domain Specification: Establish the feasible physical domain $x \in [a, b]$ or $(a, b)$.
  5. Extremum Determination: Compute $f'(x) = 0$. Use the Closed Interval Method or Second Derivative Test to verify that the critical point is indeed the global optimum.

2. The Systematic Related Rates Protocol

When physical variables are related by an equation $F(x, y, z) = 0$ and change over time $t$:

  1. Identify the given rates (e.g. $\frac{dx}{dt}$) and the target rate (e.g. $\frac{dy}{dt}$) at a specific instant $t_0$.
  2. Formulate a geometric equation relating the static variables (never substitute numerical values that change with time before differentiating!).
  3. Differentiate implicitly with respect to time $t$ using the Chain Rule: $\frac{d}{dt}[y^2] = 2y \frac{dy}{dt}$.
  4. Substitute the instantaneous values and solve algebraically for the desired rate of change.
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 8.1: Complete Critical Point and Inflection Point Determination

For the polynomial function $f(x) = x^4 - 4x^3 + 10$: (a) Find all critical numbers. (b) Determine the intervals of increase and decrease. (c) Classify all local extrema. (d) Find all points of inflection and intervals of concavity.

Step 1: Compute First Derivative and Critical Numbers
$$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) = 0 \implies x = 0, \quad x = 3$$

Critical numbers occur at $x = 0$ and $x = 3$.

Step 2: Sign Analysis of $f'(x)$
$$\begin{array}{c|c|c|c} \text{Interval} & (-\infty, 0) & (0, 3) & (3, \infty) \\ \hline 4x^2 & + & + & + \\ x - 3 & - & - & + \\ \hline f'(x) & - & - & + \end{array}$$

$f$ is strictly decreasing on $(-\infty, 3)$ and strictly increasing on $(3, \infty)$. Notice $f'$ does NOT change sign across $x = 0$.

Step 3: Classify Local Extrema
$$\text{At } x = 3: \quad f(3) = 3^4 - 4(3^3) + 10 = 81 - 108 + 10 = -17 \quad (\text{Local \& Absolute Minimum})$$

At $x = 0$, $f'$ changes from negative to negative, so $(0, 10)$ is a stationary horizontal inflection point, NOT an extremum.

Step 4: Second Derivative and Concavity
$$f''(x) = 12x^2 - 24x = 12x(x - 2) = 0 \implies x = 0, \quad x = 2$$

$f''(x) > 0$ on $(-\infty, 0) \cup (2, \infty)$ (Concave Up). $f''(x) < 0$ on $(0, 2)$ (Concave Down). Inflection points at $(0, 10)$ and $(2, -6)$.

Final Answer & Physical Insight

\text{Local/Abs Min at } (3, -17); \quad \text{Inflection Points at } (0, 10) \text{ and } (2, -6); \quad \text{Concave Up on } (-\infty, 0) \cup (2, \infty)

Tier 2: Intermediate Exam Example 8.2: Applied Geometric Optimization: Maximum Volume Cylindrical Can

A manufacturer wishes to design a closed cylindrical can of volume $V = 1000\pi\text{ cm}^3$ using the minimum possible surface area of metal. (a) Formulate the total surface area $A$ as a function of the radius $r$. (b) Determine the radius $r$ and height $h$ that minimize the surface area. (c) Prove that for any optimum cylinder, the height must equal the diameter ($h = 2r$).

Step 1: Constraint Equation and Objective Function
$$V = \pi r^2 h = 1000\pi \implies h = \frac{1000}{r^2}$$

Total surface area is two circular ends plus the cylindrical lateral mantle: $A(r) = 2\pi r^2 + 2\pi r h$.

Step 2: Express Area as a Function of $r$
$$A(r) = 2\pi r^2 + 2\pi r \left(\frac{1000}{r^2}\right) = 2\pi r^2 + \frac{2000\pi}{r} \quad (r > 0)$$

This expresses the objective function purely in terms of radius $r$.

Step 3: Differentiate and Find Critical Radius
$$A'(r) = 4\pi r - \frac{2000\pi}{r^2} = 0 \implies 4\pi r^3 = 2000\pi \implies r^3 = 500 \implies r = \sqrt[3]{500} = 5\sqrt[3]{4}\text{ cm}$$

Setting $A'(r) = 0$ gives the unique positive critical radius.

Step 4: Verify Minimum via Second Derivative Test & Height Ratio
$$A''(r) = 4\pi + \frac{4000\pi}{r^3} \implies A''(\sqrt[3]{500}) = 4\pi + \frac{4000\pi}{500} = 4\pi + 8\pi = 12\pi > 0$$

Since $A''(r) > 0$, the critical radius guarantees a strict global minimum. Computing $h$: $h = \frac{V}{\pi r^2} = \frac{\pi r^2 h}{\pi r^2} \implies h = \frac{2000}{2r^2} = 2 \left(\frac{1000}{2r^2}\right) = 2r$. Thus $h = 2r = 10\sqrt[3]{4} ext{ cm}$.

Final Answer & Physical Insight

r = 5\sqrt[3]{4} \approx 7.94\text{ cm}, \qquad h = 10\sqrt[3]{4} \approx 15.87\text{ cm}, \qquad h = 2r \text{ (Height equals Diameter)}

Tier 3: Honors / Proof Challenge Example 8.3: Related Rates Analysis: Inverted Conical Tank Drainage & Surface Recession Speed

An inverted conical water tank has a height of $H = 6.0\text{ m}$ and a top radius of $R = 2.0\text{ m}$. Water leaks out of a hole in the bottom vertex at a constant rate of $k = 0.50\text{ m}^3/\text{min}$. (a) Find a formula relating the water volume $V$ directly to the instantaneous water depth $h$. (b) Derive the rate at which the water depth is falling $\frac{dh}{dt}$ as a function of depth $h$. (c) Compute $\frac{dh}{dt}$ when the water is $h = 3.0\text{ m}$ deep and when $h = 1.0\text{ m}$ deep, explaining why the water level drops with accelerating speed as it empties.

Step 1: Relate Radius and Height by Similar Triangles
$$\frac{r}{h} = \frac{R}{H} = \frac{2.0}{6.0} = \frac{1}{3} \implies r = \frac{1}{3}h$$

Because the cone has straight cross-sectional walls, the water radius $r$ scales strictly proportionally with depth $h$.

Step 2: Express Volume as a Function of Depth $h$
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{1}{3}h\right)^2 h = \frac{\pi}{27} h^3$$

Substituting $r = h/3$ yields $V(h)$ purely in terms of depth.

Step 3: Differentiate with Respect to Time $t$
$$\frac{dV}{dt} = \frac{\pi}{27} \left(3 h^2 \frac{dh}{dt}\right) = \frac{\pi h^2}{9} \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{9}{\pi h^2} \frac{dV}{dt}$$

Given that water is leaking out at $0.50 ext{ m}^3/ ext{min}$, we have $\frac{dV}{dt} = -0.50 = -1/2 ext{ m}^3/ ext{min}$.

Step 4: Compute Rates at Specified Depths
$$\frac{dh}{dt} = -\frac{9}{2\pi h^2} \\ \text{At } h = 3\text{ m}: \quad \frac{dh}{dt} = -\frac{9}{2\pi(9)} = -\frac{1}{2\pi} \approx -0.159\text{ m/min} \\ \text{At } h = 1\text{ m}: \quad \frac{dh}{dt} = -\frac{9}{2\pi(1)} = -\frac{9}{2\pi} \approx -1.432\text{ m/min}$$

As the tank empties, the cross-sectional surface area $\pi r^2 \propto h^2$ shrinks quadratically. To sustain the constant volumetric drain rate, the water depth recession rate $\left|\frac{dh}{dt}\right| \propto 1/h^2$ increases inversely with the square of depth, accelerating as $h o 0$.

Final Answer & Physical Insight

\frac{dh}{dt} = -\frac{9}{2\pi h^2}; \quad \text{At } h = 3\text{ m}: -\frac{1}{2\pi} \approx -0.16\text{ m/min}; \quad \text{At } h = 1\text{ m}: -\frac{9}{2\pi} \approx -1.43\text{ m/min}