Reading f, f′ and f″ from One Another
- Translate features of a graph of f′ into statements about the graph of f
- Distinguish "f′ is negative" from "f′ is decreasing" and say what each implies about f
- Answer questions about f when only the graph of f′ is given
The exam usually gives you the derivative
A recurring free-response setup shows the graph of f′ and asks about f. Every question is then a translation: where f′ crosses zero going down, f has a relative maximum; where f′ is above the axis, f is increasing; where f′ itself is increasing, f is concave up; where f′ has a local extremum, f has an inflection point. Nothing new is being tested — but reading the wrong graph as f is the single most costly error in this unit, so it is worth labeling the axes out loud before answering anything.
Negative versus decreasing
These describe different graphs and get conflated constantly. "f′ is negative" means f is going down. "f′ is decreasing" means f″ < 0, so f is concave down. A function can be increasing and concave down at once — f′ positive but decreasing — which is what a quantity that is still growing but leveling off looks like. Whenever a prompt uses one of these phrases, ask which derivative the adjective belongs to.
The three-graph table
Fix the relationships once and the whole unit follows: f increasing ⟺ f′ > 0; f has a relative extremum ⟺ f′ changes sign; f concave up ⟺ f′ increasing ⟺ f″ > 0; f has an inflection point ⟺ f′ has a relative extremum ⟺ f″ changes sign. Read down the column for whichever graph you were handed. Given a graph of f′, "f′ has a minimum at x = 4" and "f has an inflection point at x = 4" are the same statement.
The graph of f′ is a line passing through (0, −4) with slope 2, on the interval [0, 5]. Describe the behavior of f.
- 1.From the description, f′(x) = 2x − 4.
- 2.f′ < 0 for x < 2 and f′ > 0 for x > 2, so f is decreasing on (0, 2) and increasing on (2, 5).
- 3.f′ changes from negative to positive at x = 2, so f has a relative minimum at x = 2.
- 4.f′ is a line of positive slope, so f′ is increasing on the whole interval — meaning f″ = 2 > 0 throughout.
- 5.f is therefore concave up on all of (0, 5), with no inflection point, since f′ has no relative extremum on the interval.
On a graph of f′, the y-intercept is f′(0), not f(0). Nothing on that graph gives you a value of f — only an initial condition plus an integral can. Questions that ask for f(3) always supply one.
The graph of f′ is positive and decreasing on (1, 6). What is true of f on that interval?
The graph of f′ has a relative maximum at x = 4. What does f have at x = 4?
When the stem says "the graph of f′ is shown," write "f′" in the margin next to the picture before reading the questions. Half the lost points in this unit come from answering as though the picture were f.
Answer the 2 checkpoints as you read.
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