Calculus AB — reading the graph of the derivative

A large share of AP Calculus AB questions never give you the function itself. They show the graph of its derivative, or a table of its values, and ask what that says about the original function: where it rises and falls, where it peaks, where its graph bends, and how high it gets. This material trains exactly that translation, the analytical applications of differentiation in Unit 5 combined with the accumulation ideas of Units 6 and 8.

The twelve quiz questions work from two drawn derivative graphs and one table. The first graph is made of straight segments, so every reading can be checked exactly: you find the relative maxima of f where f′ changes from positive to negative, pick the one complete justification for a relative minimum out of four statements that sound plausible, find the interval where f is concave up because f′ is rising, locate the inflection points at the turning points of f′, and intersect two conditions to find where f is decreasing and concave up at the same time. The same graph then asks for values: f(4) from f(0) plus the signed area under f′, and the absolute maximum of f on the closed interval, found by comparing the endpoints with every critical point.

The second graph is a smooth curve, f′(x) = (x + 1)(x − 2)², chosen because it holds the two classic traps. At x = 2 the derivative touches zero without changing sign, so there is no extremum there. At x = 0 the derivative reaches its peak, which marks the place where f climbs fastest and changes concavity, not a maximum of f; one question has you correct a student who says otherwise. A last question gives only five values of a continuous derivative and asks which conclusion is guaranteed, the kind of reasoning a table-based free-response part demands.

Every wrong option is built from a real confusion: reading the sign of f′ as concavity, calling a zero of f′ an extremum without checking the sign change, counting area below the axis as positive. The explanation after each answer shows the reasoning in the words a grader expects.

The flashcards collect the rules that link f′ and f″ to the shape of f: increasing and decreasing, the first derivative test, concavity, inflection points, the Fundamental Theorem written as f(b) = f(a) plus the integral of f′, critical points, and the candidates test for an absolute maximum.

This is independent practice written by Zestly, based on the published AP Calculus AB course framework; it is not produced or endorsed by the College Board.

  • Locate relative maxima and minima of f from the sign changes of a graphed derivative
  • Write the complete justification for a relative extremum
  • Find intervals of concavity and inflection points of f from where f′ increases, decreases and turns
  • Combine monotonicity and concavity conditions on one interval
  • Compute values of f from an initial value plus the signed area under f′
  • Find the absolute maximum of f on a closed interval by comparing endpoints and critical points
  • Draw only the conclusions a table of a continuous derivative guarantees

Practice material written by Zestly, based on the College Board AP Calculus AB course framework (Unit 5: Analytical Applications of Differentiation; Unit 6: Integration and Accumulation of Change; Unit 8: Applications of Integration), 2026–27 course and exam description.

Sample question

Graph of y = f′(x) for 0 ≤ x ≤ 7, made of straight segments joining the points (0, 2), (2, −2), (5, 1) and (7, −1). The graph crosses the x-axis at x = 1, x = 4 and x = 6.

The function $f$ is differentiable on $[0, 7]$, and the graph of its derivative $f'$ consists of straight segments joining $(0, 2)$, $(2, -2)$, $(5, 1)$ and $(7, -1)$, as drawn. At which values of $x$ in the open interval $(0, 7)$ does $f$ have a relative maximum?

See the answer

$x = 1$ and $x = 6$

A relative maximum of $f$ occurs where $f'$ changes from positive to negative. The graph of $f'$ goes from above the axis to below it at $x = 1$ and at $x = 6$. The value $x = 5$ is where $f'$ itself is highest, which tells you where $f$ rises fastest, not where $f$ peaks; at $x = 4$ the sign change is from negative to positive, a relative minimum.

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