Analytical Applications of Differentiation

Unit Overview

This unit uses the first and second derivatives to describe the shape of a graph: where it rises and falls, where it peaks and dips, and how it bends. These tools let you locate extrema, identify concavity and inflection points, and solve optimization problems ("find the maximum/minimum"). It also formalizes the key existence theorems — the Mean Value Theorem (MVT) and Extreme Value Theorem (EVT).

This is the largest analytic unit (14 topics) and a major exam contributor, especially on free-response, where justification using derivative signs is essential.

Big picture: f' controls increasing/decreasing and extrema; f'' controls concavity and inflection.

Learning Objectives

• State and apply the Mean Value Theorem and Extreme Value Theorem.
• Use f' to find intervals of increase/decrease and local extrema (First Derivative Test).
• Use f'' to determine concavity and inflection points, and apply the Second Derivative Test.
• Find absolute extrema on a closed interval (candidates test).
• Interpret relationships among the graphs of f, f', f''.
• Solve optimization problems.
• Analyze implicitly defined curves (tangents, extrema).

Prerequisite Concepts

• Fluent differentiation (Units 2–3) and critical-point solving.
• Sign analysis (number lines / sign charts).
• The theorems of Unit 1 (IVT) and continuity/differentiability.

Key Definitions

• Critical number: a value c where f'(c) = 0 or f'(c) is undefined (and f is defined at c).
• Local (relative) maximum/minimum: highest/lowest value in a neighborhood.
• Absolute (global) extremum: highest/lowest value over the entire domain/interval.
• Concave up: f'' > 0 (slopes increasing, graph opens upward); concave down: f'' < 0.
• Inflection point: a point where concavity changes (f'' changes sign).

Important Theorems

Extreme Value Theorem (EVT): A function continuous on a closed interval [a, b] attains an absolute max and an absolute min on [a, b].

Mean Value Theorem (MVT): If f is continuous on [a, b] and differentiable on (a, b), there exists c in (a, b) with f'(c) = (f(b) − f(a)) / (b − a) (a tangent parallel to the secant). Rolle's Theorem is the case f(a) = f(b), giving f'(c) = 0.

First Derivative Test: At a critical number c, if f' changes + to −, local max; − to +, local min; no sign change, neither.

Second Derivative Test: At a critical number c with f'(c) = 0: if f''(c) > 0, local min; if f''(c) < 0, local max; if f''(c) = 0, inconclusive (use the First Derivative Test).

Concept Explanations

Increasing/decreasing and extrema (5.3–5.4). Build a sign chart of f' using critical numbers. f rises where f' > 0, falls where f' < 0; sign changes locate local extrema.

Concavity and inflection (5.8, 5.10). A sign chart of f'' shows concavity. An inflection point requires f'' to change sign — a zero of f'' alone is not enough (e.g., x⁴ at 0).

Reading f, f', f'' together (5.9). These are the highest-value skills:

• f increasing ⇔ f' > 0; f has local max where f' changes + → −.
• f concave up ⇔ f'' > 0 ⇔ f' increasing.
• A local min of f' corresponds to an inflection point of f.

Absolute extrema (5.5). On a closed interval, the max/min is always among the critical values and the endpoints — just compare the candidate values.

Optimization (5.12). The art is translating words into an objective function of one variable. Classic results: for fixed perimeter the max-area rectangle is a square; for a fixed-volume closed cylinder, minimum surface area occurs when height = diameter.

MVT applications (5.6). MVT connects average and instantaneous rates: if a car averages 60 mph, at some instant its speedometer read exactly 60. It also proves that f' = 0 everywhere ⇒ f is constant, and bounds like f'(x) ≤ 2 ⇒ f(b) − f(a) ≤ 2(b − a).

Worked Examples

Example 1 (First Derivative Test). For f(x) = x³ − 3x, f'(x) = 3(x−1)(x+1). f' changes + → − at x = −1 (local max) and − → + at x = 1 (local min).

Example 2 (Second Derivative Test). Same f: f''(x) = 6x. f''(1) = 6 > 0 ⇒ local min at x = 1; f''(−1) = −6 < 0 ⇒ local max at x = −1.

Example 3 (Absolute extrema). f(x) = x³ − 3x on [0, 2]: critical x = 1. f(0) = 0, f(1) = −2, f(2) = 2. Absolute max 2 (x = 2), absolute min −2 (x = 1).

Example 4 (Inflection). f(x) = x³ − 6x²: f''(x) = 6x − 12 = 0 at x = 2, and f'' changes sign there — inflection point at x = 2.

Example 5 (Optimization). Two nonnegative numbers sum to 12; maximize the product. P = x(12 − x), P'(x) = 12 − 2x = 0 at x = 6, product = 36.

Example 6 (MVT). For f(x) = x² on [1, 3]: average rate = (9 − 1)/2 = 4; set 2c = 4 ⇒ c = 2.

Graphical Interpretation

• f' graph: above the axis ⇒ f increasing; zeros with sign change ⇒ f's local extrema; f' increasing ⇒ f concave up.
• f'' graph: sign gives concavity; sign changes give inflection points of f.
• A smooth "peak" of f sits above a downward zero-crossing of f'; a "valley" above an upward crossing.

Common Mistakes

• Calling every zero of f'' an inflection point — the sign must change.
• Reporting a critical number as the answer instead of the max/min value (or point).
• Skipping endpoints in a closed-interval absolute-extrema problem.
• Trusting the Second Derivative Test when f''(c) = 0 — it's inconclusive; use the First Derivative Test.
• Weak justification on free-response — you must cite the sign behavior of f' or f''.
• Ignoring MVT/EVT hypotheses (continuity/differentiability) when the theorem is invoked.

AP Exam Tips

• On free-response, justify with a sign chart or explicit sign statement ("f' changes from positive to negative, so f has a local max").
• Absolute extrema on [a, b]: always run the candidates test (criticals + endpoints).
• For "f is increasing/concave up on…", answer with intervals and the derivative reasoning.
• Know the classic optimization results to sanity-check answers.
• State the hypotheses when citing MVT or EVT — it earns the justification point.

Unit Summary

• f' : increasing/decreasing and local extrema (First Derivative Test).
• f'' : concavity and inflection points (Second Derivative Test, when f''(c) ≠ 0).
• Absolute extrema on [a, b] come from the candidates test (criticals + endpoints).
• MVT: f'(c) = (f(b) − f(a))/(b − a); EVT: continuous on closed interval ⇒ attains max and min.
• Optimization: reduce to one variable, then use critical-point analysis.

Next unit: the reverse process — integration and accumulation of change.