AP Calculus AB and BC · Exam guide
AP Calculus Graphical Analysis FRQ: Reading the Graph of a Derivative
In a graphical analysis question you are given the graph of a derivative, \(f'\), and asked about \(f\). Every answer comes from three facts: the sign of \(f'\), the slope of \(f'\), and the area under \(f'\). This guide translates the graph into those facts.
The three translations
| If you see on the graph of f' | Then f has | Because |
|---|---|---|
| \(f'\) above the axis | \(f\) increasing | \(f' \gt 0\) |
| \(f'\) below the axis | \(f\) decreasing | \(f' \lt 0\) |
| \(f'\) crosses from positive to negative | A relative maximum | \(f'\) changes sign |
| \(f'\) increasing (rising) | \(f\) concave up | \(f'' \gt 0\) |
| \(f'\) decreasing (falling) | \(f\) concave down | \(f'' \lt 0\) |
| Area between \(f'\) and the axis | A change in the value of \(f\) | Fundamental Theorem of Calculus |
A worked example
The graph of \(f'\) on \([0, 6]\) is made of line segments that connect the points \((0, 4)\), \((2, 0)\), \((4, -2)\), and \((6, -2)\). It is given that \(f(0) = 1\). (a) Find \(f(2)\), \(f(4)\), and \(f(6)\). (b) On what interval is \(f\) concave down? (c) At what value of \(x\) does \(f\) have its absolute maximum? Justify your answer.
(a) \(f(2) = 1 + \int_0^2 f'(x)\,dx = 1 + \dfrac{1}{2}(2)(4) = 5\). \(f(4) = 5 + \int_2^4 f'(x)\,dx = 5 - \dfrac{1}{2}(2)(2) = 3\). \(f(6) = 3 + (2)(-2) = -1\).
(b) \(f'\) is decreasing on \((0, 4)\), so \(f'' \lt 0\) and \(f\) is concave down on \((0, 4)\). On \((4, 6)\), \(f'\) is constant, so \(f'' = 0\).
(c) \(f'\) is positive on \([0, 2)\) and negative on \((2, 6]\), so \(f\) increases and then decreases, with a relative maximum at \(x = 2\). The candidates are \(f(0) = 1\), \(f(2) = 5\), and \(f(6) = -1\). The absolute maximum is at \(x = 2\).
It confuses a zero of f' with a maximum. A zero of f' is a candidate, not a proof.
It cites the sign change and checks the endpoints.
Careful wording
- Say "f'" or "f" every time. A common error is to describe the graph of \(f'\) as if it were \(f\).
- Concavity of \(f\) is about \(f''\), the slope of the graph of \(f'\), not the sign of \(f'\).
- Areas below the axis are negative contributions to a change in \(f\).
- An inflection point needs \(f''\) to change sign, which is when the graph of \(f'\) changes from rising to falling or the reverse.
Common mistakes
- Reading the graph of \(f'\) as the graph of \(f\).
- Forgetting the starting value \(f(0)\) when finding \(f\) at another point.
- Counting an area below the axis as positive.
- Skipping the endpoints in an absolute extremum question.
Frequently asked questions
Why do I need f(0)?
The area under f' gives only a change in f, so you need a starting value to find f itself.
How do I find where f has an inflection point?
Look for where f' changes from increasing to decreasing, or the reverse.
What if the graph has a corner?
At a corner of f', the function f' is not differentiable, so f'' does not exist there, but f may still have an inflection point if the concavity changes.
Is this the same for AB and BC?
Yes.
Where can I find the official scoring rules?
Exam formats change a little from year to year, so check the current course and exam description on the College Board site before test day.
Keep practicing
Every lesson in the AP Calculus courses ends with practice problems and an Exam focus box that shows how the topic is tested, and every unit has a 10-problem practice set with worked answers. These free lessons also build the skills the free-response section tests:
See the AP Calculus AB and BC course Free lessons and lesson list See the AP Calculus BC course
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Multiple choice
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