Analytical Applications of Differentiation

What this unit covers

The topics below follow the published Calculus BC course framework for Unit 5. This unit is worth 10–15% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• The Mean Value Theorem13 min · 3 objectivesState the hypotheses and conclusion of the Mean Value Theorem · Find the guaranteed value c for a given function and interval · Interpret the MVT as an average rate equaling an instantaneous rate
• Critical Points and the First Derivative Test14 min · 3 objectivesLocate critical points, including where f′ is undefined rather than zero · Use a sign chart for f′ to classify each critical point as a relative maximum, minimum, or neither · Write a justification that cites the sign change of f′ rather than the value of f
• Concavity, Inflection Points and the Second Derivative Test14 min · 3 objectivesDetermine intervals of concavity from the sign of f″ · Identify inflection points as places where f″ changes sign, not merely where f″ = 0 · Apply the Second Derivative Test and recognize when it is inconclusive
• Reading f, f′ and f″ from One Another15 min · 3 objectivesTranslate features of a graph of f′ into statements about the graph of f · Distinguish "f′ is negative" from "f′ is decreasing" and say what each implies about f · Answer questions about f when only the graph of f′ is given
• Optimization15 min · 3 objectivesTranslate a word problem into an objective function of one variable · Use critical points and endpoint analysis to find absolute extrema · Justify that a critical point is a maximum or minimum

Formulas in Unit 5

Every term in Unit 5

All 28 terms we publish for Analytical Applications of Differentiation, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Mean Value Theorem hypotheses

Continuous on the closed interval and differentiable on the open one. Stating both is required before using the conclusion.

First derivative test

f′ changing positive to negative gives a local maximum, negative to positive a local minimum. No sign change means neither.

Second derivative test

At a critical point, f″ < 0 gives a local maximum and f″ > 0 a local minimum; f″ = 0 is inconclusive.

Point of inflection

Where concavity changes, so f″ must change SIGN. f″ = 0 alone is insufficient, as x⁴ at the origin shows.

Candidates test

On a closed interval, compare f at every critical point and both endpoints to find the absolute extrema.

Reading f′ to describe f

f increases where f′ is positive and is concave up where f′ is increasing. A minimum of f′ is an inflection point of f.

Justification language

Name the theorem or the sign change and say where. "Because f′ changes from positive to negative at x = 2" scores; "because it is the maximum" does not.

Optimization procedure

Express the quantity, eliminate a variable with the constraint, differentiate, find critical points, and justify the extremum.

Extreme Value Theorem hypotheses

Continuous on a closed, bounded interval. Both parts matter — f(x) = x on (0, 1) attains neither extremum.

Candidates Test procedure

Evaluate f — not f′ — at every critical point AND both endpoints, then compare the values.

Forgetting the endpoints

The most common way to lose an absolute-extremum question. A critical point is a candidate, not an answer.

Ties are allowed

An absolute maximum value may be attained at more than one location. The question asks for the value.

Absolute extrema on an open interval

The Candidates Test does not apply. Use a sign analysis of f′ across the ENTIRE interval, or an end-behavior argument.

Critical point definition

An interior point where f′ = 0 or f′ fails to exist. The second case is routinely overlooked.

Two-row sign chart

One row for f′ giving direction, one for f″ giving curvature. Together they fix the shape of f up to vertical position.

Sign information gives shape, not height

Functions differing by a constant have identical sign charts, which is why these questions supply one value of f.

Inflection point needs a sign change

f″(c) = 0 is not sufficient. For x⁴ the second derivative vanishes at 0 and is positive on both sides.

Inflection where f″ is undefined

Possible, provided concavity actually changes and f is continuous there.

Report both coordinates

An inflection POINT has an x and a y. A bare x answers a different question.

Second derivative test is inconclusive at zero

If f″(c) = 0, fall back on the first derivative test, which always works.

Justification names a derivative

Which derivative, what it does, where — then the conclusion. Three elements, one sentence.

f′(c) = 0 justifies nothing

It holds at maxima, at minima, and at neither. The sign CHANGE is what distinguishes them.

Because the graph turns around

Description, not justification. Restating the conclusion earns nothing however fluently it is written.

Absolute extremum justification

Show the comparison of candidate values. The comparison IS the justification.

MVT justification wording

State continuity on the closed interval, differentiability on the open interval, and the average rate of change. Then conclude.

IVT justification wording

State continuity, both endpoint values, and that the target lies between them. Then conclude a c exists.

Even power blocks a sign change

For f′ = (x − 2)(x − 5)² there is no extremum at 5, because the squared factor keeps f′ positive on both sides.

Optimization: reduce to one variable

Use the constraint to express the objective in a single variable before differentiating. Two variables cannot be optimized this way.

What examiners penalize here

• MVT justifications must state both hypotheses ("f is continuous on [a,b] and differentiable on (a,b)") before invoking the conclusion. Many free-response points hinge on that explicit verification, not on solving for c.
• Free-response justifications are graded on the sign change. Write "f′ changes from positive to negative at x = 1, so f has a relative maximum there" — the phrase "changes from … to …" is what the rubric looks for.
• The exam asks "for what values of x does the graph of f have a point of inflection? Justify." A justification that stops at "f″(x) = 0 there" is incomplete every time — say that f″ changes sign.
• When the stem says "the graph of f′ is shown," write "f′" in the margin next to the picture before reading the questions. Half the lost points in this unit come from answering as though the picture were f.
• On free response you must *justify* that your answer is a max or min — cite a sign change in f′ (first-derivative test), the sign of f′′ (second-derivative test), or the candidates comparison. An unjustified extremum loses the justification point.

Practice Calculus BC

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 10 units of AP Calculus BC

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Fundamental Properties
3. Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
4. Unit 4 · Contextual Applications of Differentiation
5. Unit 5 · Analytical Applications of Differentiation
6. Unit 6 · Integration & Accumulation of Change
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration
9. Unit 9 · Parametric, Polar & Vector-Valued Functions
10. Unit 10 · Infinite Sequences & Series