AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-9)
Unit 5 · Topic 5.9
5.9 Connecting a Function, Its First Derivative, and Its Second Derivative
This topic pulls everything together: given f, f′ or f″ as a graph, table or formula, you should be able to find and justify where f increases, decreases, has extrema, is concave up or down, and has inflection points. The most common exam version gives only the graph of f′.
Key terms
- f, f′ and f″
- graph of the derivative
- relative extrema
- inflection point
- justification
The master table
Every question about the behavior of f comes back to one of these facts:
| Behavior of f | Evidence from f′ | Evidence from f″ |
|---|---|---|
| Increasing | f′ > 0 | (not directly) |
| Decreasing | f′ < 0 | (not directly) |
| Relative max at c | f′ changes + to − | f′(c) = 0 and f″(c) < 0 |
| Relative min at c | f′ changes − to + | f′(c) = 0 and f″(c) > 0 |
| Concave up | f′ increasing | f″ > 0 |
| Concave down | f′ decreasing | f″ < 0 |
| Inflection point | f′ has a relative max or min | f″ changes sign |
When you're given the graph of f′
This is the most tested setup. Treat the picture as a graph of slopes, not of f itself.
For signs of f′ (increasing, decreasing, extrema of f), look at whether the graph is above or below the x-axis. For the behavior of f′ (concavity and inflection of f), look at whether the graph is going up or down. The slope of the f′ graph is f″.
In motion problems, the same reading applies when you're given a graph of velocity: velocity is f′ of position, so above the axis means moving right, and the slope of the velocity graph is the acceleration.
When you're given a table
A table might list values of f, f′ and f″ at a few points. A row with f′(c) = 0 and f″(c) ≠ 0 lets you use the Second Derivative Test. Values of f′ that change sign between rows suggest an extremum in between, but a table alone usually can't tell you the exact behavior between rows unless more information is given.
For example, if f′(1) = 3 and f′(4) = −2 and f′ is continuous, IVT applied to f′ says f′ = 0 somewhere between 1 and 4.
Justify every claim
On free response, an answer without a reason gets no credit. Every justification should name the function (f′ or f″) and what it does: its sign, its sign change, or whether it is increasing or decreasing. Words like “the graph goes up” are too vague; say “f′ > 0” or “f′ is increasing.”
Avoid using the word “it” for different functions. “It is positive” could mean f or f′. Name the function every time.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Reading f from the graph of f′
The graph of f′ on [−3, 5] is a smooth curve. It starts at (−3, −2), crosses the x-axis going up at x = −1, rises to a peak at (1, 3), then falls, crossing the x-axis going down at x = 3, and ends at (5, −2). (a) Where does f have relative extrema? (b) Where is f concave up? (c) Where does f have a point of inflection?
Show the solutionHide the solution
- Step 1: (a) f′ changes from negative to positive at x = −1: relative minimum of f there. f′ changes from positive to negative at x = 3: relative maximum of f there.
- Step 2: (b) f is concave up where f′ is increasing: on (−3, 1).
- Step 3: (c) f′ changes from increasing to decreasing at x = 1 (the peak of f′), so the concavity of f changes there.
Answer: (a) Relative min at x = −1 and relative max at x = 3, by sign changes of f′. (b) Concave up on (−3, 1), because f′ is increasing. (c) Inflection point at x = 1, because f′ changes from increasing to decreasing.
- Example 2
Using a table of f, f′ and f″
A twice-differentiable function has f(2) = 7, f′(2) = 0 and f″(2) = −3. What can you conclude about x = 2?
Show the solutionHide the solution
- Step 1: f′(2) = 0, so x = 2 is a critical point.
- Step 2: f″(2) < 0, so the graph is concave down there.
- Step 3: Second Derivative Test: f has a relative maximum at x = 2.
Answer: f has a relative maximum value of 7 at x = 2, because f′(2) = 0 and f″(2) < 0.
- Example 3
Trap: the peak of f′ is not a peak of f
Using the same graph of f′ from the first example (peak at (1, 3)), a student says f has a relative maximum at x = 1 because the graph peaks there. What's wrong?
Show the solutionHide the solution
- Step 1: The graph shown is f′, not f. Its peak means f′ is largest at x = 1, so f is rising most steeply there.
- Step 2: f′(1) = 3 > 0, so f is increasing through x = 1. It can't have a max there.
- Step 3: A peak of f′ marks an inflection point of f.
Answer: x = 1 is an inflection point of f, not a maximum. f's relative max is at x = 3, where f′ changes from positive to negative.
Common mistakes
- Reading the graph of f′ as if it were the graph of f.
- Using x-intercepts of f′ to find inflection points of f. Use peaks and valleys of f′.
- Writing vague justifications like “it changes direction.” Name f′ or f″ and say what it does.
On the exam
- Expect at least one free-response question built on a graph of f′ (often with f defined by an integral in later units). The questions about extrema, concavity and inflection use this topic's reasoning every time.
- Use precise wording: “f′ changes from positive to negative,” “f′ is decreasing,” “f″ changes sign.”
Connected topics
Videos
Check yourself
3 questions on 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative. Pick an answer to see if you got it, and why.
The function f is continuous on the closed interval [−4, 6] and differentiable on the open interval (−4, 6). The graph of f′, the derivative of f, consists of three line segments connecting the points (−4, 2), (−2, −2), (2, 2) and (6, −2).
Described graph of f′
At which values of x in the open interval (−4, 6) does f have a relative maximum?
On which of the following intervals is f both decreasing and concave up?
At which values of x does the graph of f have a point of inflection?
0 of 3 answered