Analytical Applications of Differentiation: AP Calc AB Unit 5

Why this unit matters

Derivative tests turn symbolic expressions into decisions about increase, decrease, extrema, and shape.

What you will learn

Find intervals of increase and decrease using first-derivative sign analysis.
Classify local extrema and concavity with derivative tests.
Solve optimization problems with clear constraints and interpretations.

Understand the core ideas

Analytical applications convert derivative information into decisions about a function's behavior. Critical points occur where f'(x) = 0 or f'(x) is undefined, as long as x is in the domain of f. These points are candidates for local maxima, local minima, or neither. The first derivative test checks sign changes of f' around each critical point. If f' changes from positive to negative, the function rises then falls, so there is a local maximum. If f' changes from negative to positive, there is a local minimum.

Second derivative ideas describe shape. If f''(x) > 0, the graph is concave up; if f''(x) < 0, concave down. A sign change in f'' can indicate an inflection point when f is continuous there. Optimization adds context: define an objective quantity to maximize or minimize, rewrite it in one variable using constraints, then analyze critical points and endpoints. In AP Calculus AB, correct setup is usually the hardest step. Good practice includes labeling units, checking domain restrictions, and interpreting the numerical answer in words after calculus steps are complete.

Key terms

Critical point: A domain value where f'(x) = 0 or f'(x) does not exist, used as a candidate for extrema.
First derivative test: A sign-change test on f' around a critical point to classify local maxima or minima.
Concavity: The direction of bending of a graph, determined by the sign of the second derivative.
Optimization: Using calculus to find maximum or minimum values of a quantity subject to constraints.

Optimize rectangle area with fixed perimeter

A rectangle has perimeter 100 m. Find dimensions that maximize area.

Let sides be x and y. From 2x + 2y = 100, solve y = 50 - x.
Area is $A(x) = x(50 - x) = 50x - x^2$ , with domain 0 < x < 50.
Differentiate: A'(x) = 50 - 2x. Set A'(x) = 0 to get x = 25.
Then y = 25 and A(25) = 625. Since A is a downward-opening quadratic, this is the maximum.

Result: The maximum-area rectangle is 25 m by 25 m, with area

625 m^2

A common misconception

Claim: Any point where f''(x) = 0 is automatically an inflection point.

Correction: f''(x) = 0 is only a candidate. Concavity must actually change sign across the point.

Lessons in this unit

Critical points and first derivative testLocate and classify potential extrema from derivative behavior.
Concavity and second derivative testIdentify concavity changes and possible inflection points.
Optimization from contextWrite objective and constraint equations before differentiating. Read the full guide →
Curve sketching with derivative evidenceCombine intercepts, asymptotes, monotonicity, and concavity into one graph story.

Study task

Design an open-top box from a rectangular sheet by cutting equal corner squares. Build and optimize the volume model.

Unit checkpoint

For

f(x) = x^3 - 3x

, where are the critical points and what type are they?

$f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ , so critical points are x = -1 and x = 1. At x = -1, f changes from increasing to decreasing (local max). At x = 1, f changes from decreasing to increasing (local min).

Analytical Applications of Differentiation