Analytical Applications

What this unit covers

The topics below follow the published Calculus AB course framework for Unit 5. This unit is worth 15–18% 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

Formulas in Unit 5

Every term in Unit 5

All 17 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.

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.

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