AP Syllabus focus:
‘Intervals where f increases, decreases, or changes concavity must be consistent with the signs and shapes of both f′ and f″ when we analyze all three graphs together.’
Understanding how a function behaves requires interpreting f, f′, and f″ together, ensuring every feature matches across graphs representing values, slopes, and concavity within a given interval.
Understanding Consistency Across f, f′, and f″
Analyzing a function’s behavior demands that the graphical information from the original function, its first derivative, and its second derivative tell a coherent story. Each representation describes a different aspect of the function, and students must verify that these perspectives align. This cross-checking process is essential for interpreting problems that provide multiple representations, such as graphs, tables, and derivative sketches.
Connecting Behavior of f to Signs of f′
A primary relationship occurs between the function f and its first derivative f′, because f′ describes the instantaneous rate of change. The sign and behavior of f′ must correspond to clear features on the graph of f.
First Derivative: The derivative gives the slope of the tangent line to the graph of at each point .
When f′(x) > 0, the function must be increasing, meaning the graph of f rises as increases. When f′(x) < 0, the function must be decreasing, meaning the graph of f falls as increases. If the function displays a mismatch—such as an interval where f appears to rise while f′ is negative—then the graphs are inconsistent, signaling an error in interpretation or construction.
Connecting f′ to f″ Through Concavity
The second derivative deepens the analysis by describing how the slope itself changes across an interval. Concavity shapes the curve visually and must be consistent with what the slope graph shows.
Second Derivative: The derivative describes the rate of change of and indicates concavity of the graph of .
A region where f″(x) > 0 corresponds to concave up behavior on the graph of f, where the curve bends upward.
This figure shows a concave upward graph of y=f(x)y=f(x)y=f(x) with two tangent lines whose slopes increase from left to right. As the tangents get steeper, f′(x)f'(x)f′(x) is increasing, corresponding to f′′(x)>0f''(x) > 0f′′(x)>0. The image reinforces the relationship between tangent-line slopes and concavity. Source.
A region where f″(x) < 0 corresponds to concave down behavior, where the curve bends downward.
This figure shows a concave downward graph of y=f(x)y=f(x)y=f(x) with tangent lines whose slopes decrease from left to right. As the tangents become less steep, f′(x)f'(x)f′(x) is decreasing, reflecting that f′′(x)<0f''(x) < 0f′′(x)<0. The image highlights the direct contrast with concave-up behavior. Source.
Meanwhile, the graph of f′ should show increasing values when f″ > 0 and decreasing values when f″ < 0. These features must match; otherwise, the graphs are not aligned with the relationships dictated by differentiation.
Identifying Extrema Consistently
Local maxima and minima of f are visible as peaks or troughs on the function’s graph, but their presence must also be reflected in both derivative graphs.
Key consistency requirements include:
At a local maximum or minimum of f, the graph of f′ must cross or touch the x-axis at the corresponding -value.
The sign of f′ must change appropriately around these points.
The sign of f″ near the extremum must match the curve’s concavity.
Critical Point: A point where or where does not exist, provided the point is in the domain of .
A critical point that forms a local maximum must align with a sign change of f′ from positive to negative and with f″(x) < 0 nearby. A local minimum must align with the opposite sign change and f″(x) > 0 nearby. Any inconsistency across the three graphs indicates that at least one representation is incorrect.
Recognizing Points of Inflection Across All Graphs
Changes in concavity play a major role in shaping the function's behavior. A point where concavity switches must appear consistently across all three graphs.
A point of inflection requires:
A change in the sign of f″.
A visible shift in the bending of f.
A change in the trend of f′, meaning increasing slopes shift to decreasing slopes or vice versa.
Point of Inflection: A point where the graph of changes concavity, provided the function is defined near that point.
These features together ensure all three graphs communicate the same behavior.
Checklist for Ensuring Consistent Behavior
Students can verify alignment across f, f′, and f″ by examining each graph systematically.
This diagram summarizes how the signs of f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x) determine whether f(x)f(x)f(x) is increasing or decreasing and concave up or down. Each quadrant represents one consistent combination of slope and concavity. The content directly supports the relationships described in the study notes without adding extra topics. Source.
Use the following checks:
Increasing/Decreasing Behavior:
If f rises, f′ must be positive.
If f falls, f′ must be negative.
Extrema:
Peaks and troughs on f occur where f′ = 0.
The sign of f′ must change.
Concavity must match the extremum type.
Concavity:
If f bends upward, f′ must be increasing and f″ > 0.
If f bends downward, f′ must be decreasing and f″ < 0.
Inflection Points:
Concavity must change on f.
f′ must switch from increasing to decreasing or the reverse.
f″ must change sign.
Each representation reinforces the others, and only by checking all three can students confidently describe the full behavior of the function.
FAQ
Check for consistency in three areas:
• The sign of f' must match whether f is rising or falling.
• Peaks or troughs in f must occur where f' touches or crosses the horizontal axis.
• The curvature of f (its bending) must align with whether f' is increasing or decreasing, matching the sign of f''.
If any one of these relationships fails, the graphs cannot belong to the same function.
Students often:
• Assume f' must cross the axis at every turning point of f, forgetting it may only touch the axis.
• Confuse increasing f' with increasing f; increasing f' actually relates to concavity.
• Forget to check the entire interval, not just individual key points.
Consistency must be examined globally, not just locally.
Concavity directly links f' and f''. If f' is increasing on an interval, f'' must be positive there; if f' is decreasing, f'' must be negative.
When comparing candidates, track how steepness changes in the f' graph. This pattern should match the sign of the supposed f'' graph. If the trends don’t align, the pairing is incorrect.
Yes. Concavity depends on how the slope changes, not on whether the slope becomes zero.
A function can be entirely increasing or entirely decreasing while still switching between concave up and concave down. What matters is whether the slope becomes steeper or less steep across intervals. A concavity change requires only that the behaviour of the slope reverses direction.
Useful checks include:
• Look for where f' has maxima or minima; these occur where f'' is zero.
• Identify intervals where f' increases or decreases; this gives the sign of f''.
• Examine abrupt changes in the curvature of f'; smooth regions correspond to simple concavity behaviour in f''.
These quick checks typically eliminate incorrect candidates rapidly.
Practice Questions
Question 1 (1–3 marks)
The graph of the first derivative f' of a differentiable function f is shown to be entirely above the x-axis on the interval 1 < x < 4, and entirely below the x-axis on the interval 4 < x < 7. Describe the behaviour of f on these intervals, ensuring your answer is consistent with the information from f'.
Question 1
• 1 mark: States that f is increasing on 1 < x < 4 because f' is positive.
• 1 mark: States that f is decreasing on 4 < x < 7 because f' is negative.
• 1 mark: Uses correct terminology about increasing/decreasing behaviour linked to the sign of f'.
Question 2 (4–6 marks)
The graphs of a function f, its first derivative f', and its second derivative f'' are shown.
• On 0 < x < 2, the graph of f' increases from negative to positive values.
• On 2 < x < 4, f' decreases, remaining positive.
• The graph of f'' crosses the x-axis at x = 2 and changes sign from positive to negative.
• On 4 < x < 6, the graph of f' continues decreasing and crosses the x-axis at x = 5.
Using consistent reasoning across all three graphs (f, f', and f''), answer the following:
(a) State where f is increasing or decreasing and justify each interval.
(b) Identify any local extremum of f and justify it.
(c) State whether there is a point of inflection for f and justify your answer.
(d) Describe the concavity of f on each interval.
Question 2
(a)
• 1 mark: f is increasing on 0 < x < 6 wherever f' > 0; specifically, increasing on the entire interval except where noted otherwise.
• 1 mark: Recognises that f is decreasing nowhere before x = 6, except after x = 5 where f' < 0.
• Answers must justify using the sign of f'.
(b)
• 1 mark: Identifies a local minimum of f at x = 2 because f' changes from negative to positive.
• 1 mark: Identifies a local maximum of f at x = 5 because f' changes from positive to negative.
(c)
• 1 mark: States that there is a point of inflection at x = 2 because f'' changes sign there.
(d)
• 1 mark: States f is concave up on 0 < x < 2 because f'' > 0.
• 1 mark: States f is concave down on 2 < x < 6 because f'' < 0.
