Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
Three graphs, one function, and the exam keeps giving you the wrong one on purpose. A graph of f′ tells you where f turns and where f bends, but it says nothing about how high f sits, because adding a constant to a function does not change its derivative. A graph of f shows the height directly and hides the slope in its own shape, so you have to read steepness by eye. The translation is mechanical once you name it: level of f′ gives the direction of f, slope of f′ gives the concavity of f, and area under f′ gives the change in f. Most lost points on this topic come from reading the wrong graph against the wrong question.
Four of the nine released exams since 2017 build a whole question around a graph of f′, and each one asks for a translation rather than a computation.
Figure 1 — Each row reads one graph to answer a question about another. Signs flow downward; questions climb back up.
f′′>0⟹f′ increasing⟹f concave up
The arrows in the figure point one way and the exam questions point the other. You are shown the bottom row and asked about the top, or shown the middle row and asked about the bottom. Reading a relation backwards is not the same skill as reading it forwards, which is why this topic has its own page.
The limits of each reading matter more than the readings themselves, because the exam deliberately asks for something the given graph cannot supply.
| You are given | You can read directly | You cannot read |
|---|---|---|
| the graph of f | height, rough slope, rough bend | exact slopes, exact areas |
| the graph of f′ | sign of f, zeros of f, extrema, concavity, change in f | the value of f without a starting point |
| the graph of f′′ | concavity of f, inflection candidates | the value of f, or the slope of f |
The middle row is the one that appears most often.
A graph of f′ is generous: it answers almost every question about f except the absolute one. Without a value such as f(2)=5 in the stem, no amount of area under f′ will produce a height, because the constant of integration is invisible in a derivative.
One number on the graph of f′ answers several questions at once, and the order matters.
The third item is the one students skip. A maximum of f′ is not a maximum of f; it is the location where f climbs fastest, and it is also an inflection candidate because f′′ changes sign there.
The signed area under f′ from a to b equals f(b)−f(a). That statement is the Fundamental Theorem wearing graphical clothes, and it is the only bridge from a derivative graph to a value.
f(b)=f(a)+∫abf′(x)dx
Read the right side as a starting height plus a total change. When the stem supplies f(a), the bridge is open. When it does not, the question cannot be answered, and saying so is a legitimate response on the released exams.
Borrowed idea: The three graphs are three views of the same object, and the exam keeps handing you the wrong one on purpose. A graph of f′ tells you where f turns and where f bends, but it tells you almost nothing about how high f sits, because a constant can be added to f without changing its derivative at all. A graph of f tells you the height directly and hides the slope inside its own shape, so you have to read steepness by eye. The skill is translation, not computation. Read the level when the question is about signs. Read the slope when the question is about values. Stop mixing them.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "does f have a relative maximum at x=6" | graph of f′ crossing the axis | crossing direction | report the pattern, then the type |
| "point of inflection" | graph of f′ turning around | where f′ stops increasing | name the x where the trend of f′ flips |
| "concave up on which intervals" | graph of f′ rising | rising segments of f′ | read the x-intervals where f′ climbs |
| "find f(5) given f(2)=3" | graph of f′ and one value | signed area from 2 to 5 | add the area to the starting value |
| "at what x is f increasing fastest" | graph of f′ with a peak | the maximum level of f′ | name the input, not the value |
① Name the graph you were given and the graph the question is about. Two nouns, one sentence, before any work.
② For a value question, look for the starting value in the stem. If it is absent, the question is asking about change, not about height.
③ Write the translation clause into the answer. "Because f′ is increasing on that interval" is the reason, and the reason is the point.
Core Archetype (Simplest Form): The graph of f′ consists of two line segments, from (0,2) down to (2,0) and then up to (4,4). Given f(0)=1, find f(2) and f(4), and state where f has a relative minimum.
Standard Execution:
- Read the sign. f′<0 on (0,2) and f′>0 on (2,4), so f decreases then increases and has a relative minimum at x=2.
- Areas. The signed area from 0 to 2 is −21(2)(2)=−2, and from 2 to 4 it is 21(2)(4)=4.
- Values. f(2)=1−2=−1 and f(4)=−1+4=3.
- Check. The minimum value −1 matches the sign change, which is the consistency check that catches a sign error in the areas.
Exam Variation (Context / Parameter Shift): The graph of f′ consists of three line segments joining (0,2), (2,0), (4,4), and (6,1), and f(0)=1. On what intervals is the graph of f concave up, and where does f have an inflection point?
Solution: Concavity of f is the monotonicity of f′. The segments show f′ decreasing from x=0 to x=2, then increasing from x=2 to x=4, then decreasing from x=4 to x=6. So f is concave down on (0,2), concave up on (2,4), and concave down on (4,6). The trend of f′ changes at x=2 and at x=4, so those are the inflection points of f. Notice that x=2 is both a relative minimum of f and an inflection point, which is possible and common.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f′ crosses the x-axis at x=3, is above the axis on (0,3), and is below the axis on (3,6). Which statement about f is correct?
MCQ 2
The graph of f′ is decreasing on (−2,0) and increasing on (0,3). Which statement about f must be true?
MCQ 3
The graph of f′ and the line y=0 bound a region of signed area −3 between x=1 and x=4. If f(1)=8, what is f(4)?
FRQ 4
The graph of f′, the derivative of the differentiable function f, consists of three line segments joining the points (0,2), (2,0), (4,4), and (6,1). It is known that f(0)=1.
(a) Find the values of f(2), f(4), and f(6). Show the work that leads to your answers.
(b) At what value of x does f have a relative minimum? Give a reason for your answer.
(c) On what open intervals is the graph of f concave up? On what open intervals is it concave down? Give a reason for each answer.
(d) At what value of x in the interval 0<x<6 is f increasing most rapidly? Explain your reasoning.
| Read this on the graph of f′ | Answer shape | |
|---|---|---|
| where does f turn | where f′ crosses the axis | an input, plus the type |
| how steep is f | the level of f′ | a slope, with units |
| where is f concave up | where f′ is increasing | an open interval |
| where does f bend the other way | where the trend of f′ flips | an input |
| how much did f change | signed area under f′ | a number, not a height |
| what is f(b) | starting value plus the area | a height, if a starting value is given |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
Three graphs, one function, and the exam keeps giving you the wrong one on purpose. A graph of f′ tells you where f turns and where f bends, but it says nothing about how high f sits, because adding a constant to a function does not change its derivative. A graph of f shows the height directly and hides the slope in its own shape, so you have to read steepness by eye. The translation is mechanical once you name it: level of f′ gives the direction of f, slope of f′ gives the concavity of f, and area under f′ gives the change in f. Most lost points on this topic come from reading the wrong graph against the wrong question.
Four of the nine released exams since 2017 build a whole question around a graph of f′, and each one asks for a translation rather than a computation.
Figure 1 — Each row reads one graph to answer a question about another. Signs flow downward; questions climb back up.
f′′>0⟹f′ increasing⟹f concave up
The arrows in the figure point one way and the exam questions point the other. You are shown the bottom row and asked about the top, or shown the middle row and asked about the bottom. Reading a relation backwards is not the same skill as reading it forwards, which is why this topic has its own page.
The limits of each reading matter more than the readings themselves, because the exam deliberately asks for something the given graph cannot supply.
| You are given | You can read directly | You cannot read |
|---|---|---|
| the graph of f | height, rough slope, rough bend | exact slopes, exact areas |
| the graph of f′ | sign of f, zeros of f, extrema, concavity, change in f | the value of f without a starting point |
| the graph of f′′ | concavity of f, inflection candidates | the value of f, or the slope of f |
The middle row is the one that appears most often.
A graph of f′ is generous: it answers almost every question about f except the absolute one. Without a value such as f(2)=5 in the stem, no amount of area under f′ will produce a height, because the constant of integration is invisible in a derivative.
One number on the graph of f′ answers several questions at once, and the order matters.
The third item is the one students skip. A maximum of f′ is not a maximum of f; it is the location where f climbs fastest, and it is also an inflection candidate because f′′ changes sign there.
The signed area under f′ from a to b equals f(b)−f(a). That statement is the Fundamental Theorem wearing graphical clothes, and it is the only bridge from a derivative graph to a value.
f(b)=f(a)+∫abf′(x)dx
Read the right side as a starting height plus a total change. When the stem supplies f(a), the bridge is open. When it does not, the question cannot be answered, and saying so is a legitimate response on the released exams.
Borrowed idea: The three graphs are three views of the same object, and the exam keeps handing you the wrong one on purpose. A graph of f′ tells you where f turns and where f bends, but it tells you almost nothing about how high f sits, because a constant can be added to f without changing its derivative at all. A graph of f tells you the height directly and hides the slope inside its own shape, so you have to read steepness by eye. The skill is translation, not computation. Read the level when the question is about signs. Read the slope when the question is about values. Stop mixing them.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "does f have a relative maximum at x=6" | graph of f′ crossing the axis | crossing direction | report the pattern, then the type |
| "point of inflection" | graph of f′ turning around | where f′ stops increasing | name the x where the trend of f′ flips |
| "concave up on which intervals" | graph of f′ rising | rising segments of f′ | read the x-intervals where f′ climbs |
| "find f(5) given f(2)=3" | graph of f′ and one value | signed area from 2 to 5 | add the area to the starting value |
| "at what x is f increasing fastest" | graph of f′ with a peak | the maximum level of f′ | name the input, not the value |
① Name the graph you were given and the graph the question is about. Two nouns, one sentence, before any work.
② For a value question, look for the starting value in the stem. If it is absent, the question is asking about change, not about height.
③ Write the translation clause into the answer. "Because f′ is increasing on that interval" is the reason, and the reason is the point.
Core Archetype (Simplest Form): The graph of f′ consists of two line segments, from (0,2) down to (2,0) and then up to (4,4). Given f(0)=1, find f(2) and f(4), and state where f has a relative minimum.
Standard Execution:
- Read the sign. f′<0 on (0,2) and f′>0 on (2,4), so f decreases then increases and has a relative minimum at x=2.
- Areas. The signed area from 0 to 2 is −21(2)(2)=−2, and from 2 to 4 it is 21(2)(4)=4.
- Values. f(2)=1−2=−1 and f(4)=−1+4=3.
- Check. The minimum value −1 matches the sign change, which is the consistency check that catches a sign error in the areas.
Exam Variation (Context / Parameter Shift): The graph of f′ consists of three line segments joining (0,2), (2,0), (4,4), and (6,1), and f(0)=1. On what intervals is the graph of f concave up, and where does f have an inflection point?
Solution: Concavity of f is the monotonicity of f′. The segments show f′ decreasing from x=0 to x=2, then increasing from x=2 to x=4, then decreasing from x=4 to x=6. So f is concave down on (0,2), concave up on (2,4), and concave down on (4,6). The trend of f′ changes at x=2 and at x=4, so those are the inflection points of f. Notice that x=2 is both a relative minimum of f and an inflection point, which is possible and common.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f′ crosses the x-axis at x=3, is above the axis on (0,3), and is below the axis on (3,6). Which statement about f is correct?
MCQ 2
The graph of f′ is decreasing on (−2,0) and increasing on (0,3). Which statement about f must be true?
MCQ 3
The graph of f′ and the line y=0 bound a region of signed area −3 between x=1 and x=4. If f(1)=8, what is f(4)?
FRQ 4
The graph of f′, the derivative of the differentiable function f, consists of three line segments joining the points (0,2), (2,0), (4,4), and (6,1). It is known that f(0)=1.
(a) Find the values of f(2), f(4), and f(6). Show the work that leads to your answers.
(b) At what value of x does f have a relative minimum? Give a reason for your answer.
(c) On what open intervals is the graph of f concave up? On what open intervals is it concave down? Give a reason for each answer.
(d) At what value of x in the interval 0<x<6 is f increasing most rapidly? Explain your reasoning.
| Read this on the graph of f′ | Answer shape | |
|---|---|---|
| where does f turn | where f′ crosses the axis | an input, plus the type |
| how steep is f | the level of f′ | a slope, with units |
| where is f concave up | where f′ is increasing | an open interval |
| where does f bend the other way | where the trend of f′ flips | an input |
| how much did f change | signed area under f′ | a number, not a height |
| what is f(b) | starting value plus the area | a height, if a starting value is given |
Strengthen this topic with AP-style practice, or continue to next topic.