Analytical Applications

What this unit covers

The topics below follow the published Calculus AB course framework for Unit 5. This unit is worth 15–20% 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 · Verify that a function qualifies for the MVT on an interval · Find the guaranteed point where the instantaneous rate equals the average rate
• Critical Points & the First Derivative Test14 min · 3 objectivesFind critical points where f′ = 0 or f′ is undefined · Determine intervals of increase and decrease from the sign of f′ · Classify local maxima and minima with the First Derivative Test
• Concavity & the Second Derivative Test14 min · 3 objectivesUse the sign of f″ to determine concavity · Locate inflection points where concavity changes · Classify critical points with the Second Derivative Test
• Optimization15 min · 3 objectivesTranslate a word problem into an objective function with a constraint · Reduce the objective to a single variable and find its extrema · Verify a candidate is the true maximum or minimum and check endpoints
• Absolute Extrema and the Candidates Test14 min · 3 objectivesDistinguish absolute from local extrema and state the Extreme Value Theorem · Apply the Candidates Test on a closed interval · Handle absolute extrema on an open or unbounded interval, where the test does not apply
• Reconstructing f from f′ and f″15 min · 3 objectivesTranslate sign information about f′ and f″ into shape information about f · Locate inflection points correctly, including the requirement that concavity change · Sketch a curve consistent with a given sign chart
• Justification: Writing the Sentence That Scores15 min · 3 objectivesState what a calculus justification must contain to earn credit · Write justifications for extrema, concavity, and theorem applications · Recognize common justifications that read as complete and earn nothing

Formulas in Unit 5

Every term in Unit 5

All 45 terms we publish for Analytical Applications, 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

If f is continuous on [a, b] and differentiable on (a, b), some c in (a, b) has f′(c) = [f(b) − f(a)]/(b − a). Both hypotheses must be stated.

First derivative test

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

Rolle's Theorem

The special case of the MVT with f(a) = f(b), guaranteeing a c where f′(c) = 0.

Critical point

Where f′(x) = 0 or f′ is undefined, and x is in the domain. Not every critical point is an extremum.

Second derivative test

At a critical point, f″ < 0 gives a local maximum and f″ > 0 a local minimum. Inconclusive when f″ = 0.

Concavity

f″ > 0 is concave up and f′ is increasing; f″ < 0 is concave down and f′ is decreasing.

Point of inflection

Where concavity changes, so f″ changes SIGN. f″ = 0 alone is not sufficient — x⁴ has f″(0) = 0 with no inflection.

Candidates test for absolute extrema

On a closed interval, evaluate f at every critical point and at both endpoints; the largest and smallest values are the absolute extrema.

Optimization procedure

Write the quantity to be optimized, use a constraint to reduce it to one variable, differentiate, find critical points, and justify that yours is the extremum.

Justification language

Cite the sign change of f′ or the value of f″, and say where. "Because f′ changes from positive to negative at x = 2" earns the point; "because it is the maximum" does not.

Reading f′ to describe f

Where f′ is positive f increases; where f′ is increasing f is concave up. A zero of f′ that is a minimum of f′ is an inflection point of f.

Justifying an absolute extremum

Cite the candidates test explicitly: compare values at all critical points and both endpoints, and say which is largest.

Justifying an inflection point

State that f″ changes sign at that x, and say from what to what. "f″ = 0" alone is not a justification.

Reading a graph of f′ to answer about f

f increases where f′ is above the axis; f has a maximum where f′ crosses from above to below; f is concave up where f′ is increasing.

Sketching f from f′

The zeros of f′ locate f's extrema and the extrema of f′ locate f's inflection points. The vertical position needs an initial condition.

Comparing two functions on an interval

Show one difference is positive at a point and that the difference has no zero on the interval, so the ordering cannot reverse.

Optimization justification

A critical point is not an answer. Justify with a first or second derivative test, or with the candidates test on a closed interval.

Local versus absolute

Local compares with nearby values; absolute compares across the whole interval. Different questions, different methods.

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

Find critical points, then evaluate f — not f′ — at every critical point AND both endpoints, and 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 can 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 full-interval sign analysis of f′, or an end-behavior argument.

Critical point definition

An interior point where f′ = 0 OR f′ fails to exist. Points where f′ is not defined are frequently overlooked.

Sign chart with two rows

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

Sign information gives shape, not height

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

Four shape combinations

f′ and f″ both positive: increasing, concave up. Both negative: decreasing, concave down. Mixed signs give the two levelling cases.

Inflection point requires a sign change

f″(c) = 0 is not sufficient. For x⁴, f″ = 12x² vanishes at 0 and is positive on both sides — no inflection point.

Inflection where f″ is undefined

Possible, provided concavity actually changes there and f itself is continuous at the point.

Report both coordinates

An inflection POINT needs an x and a y. A bare x value 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 must name a derivative

Which derivative, what it does (sign or sign change), and where — then the conclusion. Three elements, one sentence.

f′(3) = 0 justifies nothing

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

Because the graph turns around

A description, not a justification. Readers award nothing for restating the conclusion in different words.

Absolute extremum justification

Show the comparison: list f at the critical points and endpoints and state which is largest. The comparison IS the justification.

MVT justification wording

State continuity on the closed interval, differentiability on the open interval, and compute 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.

Rolle as a special case

The MVT with f(a) = f(b), so the average rate is zero and some c has f′(c) = 0 — a horizontal tangent.

MVT gives an open-interval c

Never an endpoint, and the theorem does not say how many such c exist or where.

Optimization: write the constraint

Express the quantity to optimize as a function of ONE variable using the constraint, then differentiate. Two variables cannot be optimized by this method.

Optimization: justify the extremum

A closed-interval domain calls for the Candidates Test; an open one calls for a first-derivative sign argument. Do not stop at the critical point.

Optimization: answer the question asked

Sometimes the dimensions are wanted, sometimes the maximum value. Reporting the wrong one loses the point after all the work.

Even power blocks a sign change

For f′ = (x − 1)(x − 4)² there is no extremum at x = 4, because the squared factor keeps f′ positive on both sides.

Increasing on an interval versus at a point

Increasing is a property of an interval. A single value f′(c) > 0 supports a statement about a neighborhood, not about the whole domain.

Comparing f at two points using f′

If f′ > 0 throughout [a, b], then f(b) > f(a). Sign of the derivative orders the values without computing them.

What examiners penalize here

• To use the MVT on the exam, first state that f is continuous on [a, b] and differentiable on (a, b), then compute the average rate and set f′(c) equal to it. Solving for c without confirming the hypotheses loses justification credit.
• A critical point is only a *candidate* for an extremum — you must show a sign change in f′ to confirm it. Where f′ = 0 but does not change sign (as at x = 0 for y = x³), there is no local max or min.
• When the Second Derivative Test returns f″(c) = 0, do not guess — state that the test is inconclusive and switch to the First Derivative Test, examining the sign change of f′ around c to classify the point.
• Optimization free-response points hinge on justification: after finding a critical point, state a reason it gives the extreme value (sign of f′ or f″), and on a closed domain compare it against the endpoint values. A number with no justification rarely earns full credit.
• The most common way to lose this question is to **forget the endpoints**. A critical point is a candidate, not an answer, and on many problems the absolute extremum sits at an endpoint where f′ is not zero at all. Write all candidates in a table before comparing.
• Justifications that earn nothing, seen constantly: "because f′ = 0 there" (true of maxima, minima and neither); "because the graph turns around" (describes rather than justifies); "because it is the highest point" (restates the conclusion). Each of these accompanies correct arithmetic and still scores zero on the justification point.

Practice Calculus AB

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 8 units of AP Calculus AB

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Rules
3. Unit 3 · Composite & Implicit Differentiation
4. Unit 4 · Contextual Applications
5. Unit 5 · Analytical Applications
6. Unit 6 · Integration & Accumulation
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration