AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-8)
Unit 5 · Topic 5.8
5.8 Sketching Graphs of Functions and Their Derivatives
The graph of f and the graph of f′ are linked feature by feature. From f you can sketch f′ by tracking slopes, and from information about f′ and f″ you can sketch the shape of f. The key is matching each feature of one graph to its partner on the other.
Key terms
- graph of f′
- horizontal tangent
- slope
- key features
- curve sketching
From the graph of f to the graph of f′
The height of f′ at each x is the slope of f there. Walk along the graph of f from left to right and track the slope:
| On the graph of f | On the graph of f′ |
|---|---|
| Horizontal tangent (peak, valley or flat spot) | f′ crosses or touches the x-axis |
| Rising | Above the x-axis |
| Falling | Below the x-axis |
| Concave up (slopes increasing) | Rising |
| Concave down (slopes decreasing) | Falling |
| Inflection point (slope at a local max or min) | Peak or valley of f′ |
| Corner or cusp | f′ jumps or is undefined (open circles) |
From f′ and f″ to the graph of f
If you know where f′ and f″ are positive and negative, you know the shape of f on each interval. Combine the two sign charts. There are four basic shapes:
- f′ > 0 and f″ > 0: rising and bending up, getting steeper.
- f′ > 0 and f″ < 0: rising and bending down, leveling off.
- f′ < 0 and f″ < 0: falling and bending down, getting steeper.
- f′ < 0 and f″ > 0: falling and bending up, leveling off.
A sketching routine
A graph of f only needs the right shape in the right places, not perfect scale.
- Plot any known points, like f(0) or the values at critical points.
- Mark critical points and inflection points on the x-axis.
- For each interval between them, pick the matching shape from the four above.
- Connect the pieces smoothly, with peaks and valleys at the extrema and bends switching at inflection points.
- Check end behavior and any asymptotes.
Matching questions
Multiple-choice questions often show the graph of f and ask which graph is f′ (or the reverse). Pick one feature, like where f has a horizontal tangent, and eliminate options that don't have f′ = 0 there. Then check where f rises and falls against the sign of f′.
Going from a graph of f′ back to the exact heights of f takes more than slopes. You'd need to know the starting value of f and how much it rises or falls on each stretch, which is the job of integrals in Unit 6. For now, the shape is enough.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Describing f′ from f
f(x) = x³ − 3x has a relative max at (−1, 2) and a relative min at (1, −2). Describe the graph of f′ without computing it, then check with the formula.
Show the solutionHide the solution
- Step 1: f has horizontal tangents at x = −1 and x = 1, so f′ is 0 there.
- Step 2: f rises on (−∞, −1), falls on (−1, 1) and rises on (1, ∞), so f′ is positive, then negative, then positive.
- Step 3: f has an inflection point at x = 0 (its steepest downhill spot), so f′ has its lowest point there.
- Step 4: Check: f′(x) = 3x² − 3 is an upward parabola with zeros at ±1 and vertex (0, −3).
Answer: f′ is an upward-opening parabola crossing the x-axis at x = −1 and x = 1, with its minimum at (0, −3).
- Example 2
Sketching f from sign information
f is continuous with f(−2) = 5 and f(3) = −4. f′ > 0 on (−∞, −2) and (3, ∞), f′ < 0 on (−2, 3). f″ < 0 on (−∞, 1) and f″ > 0 on (1, ∞). Describe the graph of f.
Show the solutionHide the solution
- Step 1: On (−∞, −2): rising and concave down. It levels off into a peak at (−2, 5), a relative max.
- Step 2: On (−2, 1): falling and concave down, getting steeper.
- Step 3: At x = 1: concavity changes from down to up, so there's an inflection point while the graph is falling.
- Step 4: On (1, 3): falling and concave up, leveling off into a valley at (3, −4), a relative min.
- Step 5: On (3, ∞): rising and concave up.
Answer: A peak at (−2, 5), a downhill stretch that is steepest at the inflection point x = 1, a valley at (3, −4), then rising and curving upward. The overall shape is like a cubic graph with a positive leading coefficient.
- Example 3
Trap: connecting a jump in f′
Sketch the graph of f′ for f(x) = |x − 2|.
Show the solutionHide the solution
- Step 1: For x < 2, f(x) = 2 − x has slope −1. For x > 2, f(x) = x − 2 has slope 1.
- Step 2: So f′(x) = −1 on the left and 1 on the right.
- Step 3: At x = 2, f has a corner, so f′(2) doesn't exist.
- Step 4: Don't draw a line connecting −1 to 1. The graph of f′ has a gap there.
Answer: Two horizontal rays: y = −1 for x < 2 and y = 1 for x > 2, each ending in an open circle at x = 2, with no point at x = 2.
Common mistakes
- Putting the extrema of f′ where f has extrema. Extrema of f match zeros of f′; extrema of f′ match inflection points of f.
- Drawing f′ connected across a corner of f.
- Confusing the height of f with its slope, such as making f′ positive wherever f is above the x-axis.
On the exam
- Graph-matching questions are common in multiple choice. Eliminate options with one clear feature at a time.
- Free-response questions rarely ask for full sketches now, but they test the same feature matching through justifications.
Connected topics
- Unit 55.3 Determining Intervals on Which a Function Is Increasing or Decreasing
- Unit 55.6 Determining Concavity of Functions over Their Domains
- Unit 55.9 Connecting a Function, Its First Derivative, and Its Second Derivative
- Unit 22.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
Videos
Check yourself
4 questions on 5.8 Sketching Graphs of Functions and Their Derivatives. Pick an answer to see if you got it, and why.
The function f is twice differentiable on the interval (0, 6). On (0, 2), f is increasing and concave down. The graph of f has a horizontal tangent at x = 2. On (2, 4), f is decreasing and concave down, and on (4, 6), f is decreasing and concave up. Which of the following is true about f′?
A function f is twice differentiable for all x. On (−∞, 0), f is increasing and concave up. On (0, 3), f is increasing and concave down. The graph of f has a horizontal tangent line at x = 3, and on (3, ∞), f is decreasing and concave down. Which of the following is true about the graph of f′ ?
Let f(x) = x³ − 3x. Which of the following describes the graph of y = f′(x) ?
A function f is twice differentiable for all real x, with f(0) = 1, f′(x) > 0 for all x and f″(x) < 0 for all x. Which of the following could be f(x) ?
0 of 4 answered