Lesson 22: Concavity, Inflection Points & the Second Derivative Test
(a) Opening Question
In Lesson 21 you used the sign of f' to decide where a function rises and falls. Now look one derivative deeper.
Consider f(x) = x³. Its derivative is f'(x) = 3x², which is never negative — so f is always increasing. And yet the graph of y = x³ clearly bends: it curves downward to the left of the origin and upward to the right.
Increasing tells you the function is going up. But how fast the slope itself is changing — whether the curve cups upward like a bowl or downward like a dome — is a separate question. The slope of y = x³ is 3x². To the left of x = 0 the slope is decreasing (from large positive, down toward 0); to the right it is increasing again.
Your task: Compute f''(x) for f(x) = x³. Where is f'' negative? Where is it positive? What happens exactly at x = 0? Hold your answer — by the end of this lesson you will be able to say precisely why the origin is the most interesting point on this curve.
(b) Core Concepts
Concavity: the sign of the second derivative
The second derivative f''(x) is the derivative of f'(x) — the rate of change of the slope. It measures how the curve bends.
Definition (Concavity). On an interval where
f''(x) > 0, the graph offis concave up (it opens upward, like a cup ∪). On an interval wheref''(x) < 0, the graph offis concave down (it opens downward, like a cap ∩).
There are two equivalent ways to think about this, and the AP exam rewards both:
- In terms of
f'':f''>0⇒ concave up;f''<0⇒ concave down. - In terms of
f': Concave up means the slopef'is increasing; concave down means the slopef'is decreasing. (Concave up = slopes climbing from steep-negative toward steep-positive. Concave down = slopes falling.)
That second framing is exactly the link to Lesson 21: concavity of f is the increasing/decreasing behavior of f'. Apply the First-Derivative-Test logic from L21 — but to f' instead of f.
A quick f'' sign chart is the standard tool. For f(x) = x³ − 3x²:
f'(x) = 3x² − 6x
f''(x) = 6x − 6 = 6(x − 1)f''(x) = 0 at x = 1. Test a point on each side:
Interval: (−∞, 1) (1, ∞)
Test x: 0 2
f''(x): 6(−1)=−6 6(1)=6
Sign: − +
Concavity: concave DOWN concave UPInflection points: where concavity changes
Definition (Inflection Point). A point on the graph of
fis an inflection point if (i) the concavity changes there — that is,f''changes sign — and (ii) the point is in the domain off(the function is actually defined and continuous there).
The candidates for inflection are the x-values where f''(x) = 0 or f''(x) is undefined. But — and this is the single most-tested subtlety in this lesson:
⚠️
f''(x) = 0is necessary but NOT sufficient. Finding wheref''is zero only gives you candidates. You must confirm thatf''actually changes sign there. If the sign is the same on both sides, there is no inflection point.
The textbook trap is f(x) = x⁴. Here f''(x) = 12x², which equals 0 at x = 0 — but 12x² is positive on both sides of 0. Concavity never changes (concave up throughout), so x = 0 is not an inflection point, despite f''(0) = 0.
For f(x) = x³ − 3x² above, f'' changes from − to + at x = 1, and x = 1 is in the domain, so there is an inflection point at x = 1. Its y-coordinate is f(1) = 1 − 3 = −2, giving the point (1, −2).
The Second Derivative Test
Concavity gives a fast way to classify critical points. If c is a critical point where the tangent is horizontal (f'(c) = 0), the local shape near c tells you which kind of extremum it is.
The Second Derivative Test. Suppose
f'(c) = 0andf''(c)exists.- If
f''(c) > 0— the curve is concave up atc(a cup) — thenfhas a relative MINIMUM atx = c.- If
f''(c) < 0— the curve is concave down atc(a cap) — thenfhas a relative MAXIMUM atx = c.- If
f''(c) = 0— the test is INCONCLUSIVE. Fall back to the First Derivative Test (L21).
The mnemonic: a horizontal tangent sitting at the bottom of a cup (f'' > 0) is a low point (min); a horizontal tangent at the top of a cap (f'' < 0) is a high point (max). Positive-second-derivative ⇒ min feels backwards to many students — anchor it to the cup.
When f''(c) = 0, the test genuinely tells you nothing. The point could be a min (x⁴ at 0), a max (−x⁴ at 0), or neither (x³ at 0 — a horizontal inflection). You must return to the sign of f' on each side.
Reading it all from a graph
The three graphs are tightly linked. The key translations to memorize:
f'increasing ⇒fconcave up. (On the graph off': wheref'rises,fcups up.)f'decreasing ⇒fconcave down.- An inflection point of
foccurs wheref'has a local max or min (wheref'turns around) — equivalently wheref''crosses zero.
![three stacked panels for f(x) = x³ − 3x² on [−1.5, 3.5] × [−5, 3] TOP panel — f(x) = x³ − 3x²:](calciq-graphs/L22-0.svg)
Notice how x = 1 lines up across all three panels: inflection of f, turning point (minimum) of f', and sign-change (zero-crossing) of f''. That alignment is the heart of "Connecting Representations."
(c) Worked Examples
Example 1 — Concavity intervals and inflection points (NO calculator)
Problem. For f(x) = x⁴ − 4x³, find all intervals of concavity and all inflection points. Justify.
Strategy. Compute f'', find where it is 0 or undefined, build a sign chart, confirm sign changes.
Solution.
f'(x) = 4x³ − 12x²
f''(x) = 12x² − 24x = 12x(x − 2)f''(x) = 0 at x = 0 and x = 2 (a polynomial, never undefined). Sign chart:
Interval: (−∞, 0) (0, 2) (2, ∞)
Test x: −1 1 3
f'': 12(−1)(−3) 12(1)(−1) 12(3)(1)
= +36 = −12 = +36
Sign: + − +
Concavity: UP DOWN UPf'' changes + → − at x = 0 and − → + at x = 2; both x-values are in the domain.
Justification. f is concave up on (−∞, 0) and (2, ∞) and concave down on (0, 2). f has inflection points at x = 0 and x = 2 because f'' changes sign at each. The points are (0, f(0)) = (0, 0) and (2, f(2)) = (2, 16 − 32) = (2, −16).
Example 2 — Second Derivative Test classification (NO calculator)
Problem. Use the Second Derivative Test to classify the critical points of f(x) = x³ − 6x² + 9x + 2.
Strategy. Find critical points (f' = 0), then evaluate f'' at each.
Solution.
f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3)
f''(x) = 6x − 12 = 6(x − 2)Critical points: x = 1 and x = 3 (f' is defined everywhere). Now test:
f''(1) = 6(1) − 12 = −6 < 0⇒ concave down ⇒ relative MAXIMUM atx = 1. Value:f(1) = 1 − 6 + 9 + 2 = 6.f''(3) = 6(3) − 12 = 6 > 0⇒ concave up ⇒ relative MINIMUM atx = 3. Value:f(3) = 27 − 54 + 27 + 2 = 2.
Justification. By the Second Derivative Test, f has a relative maximum at x = 1 because f'(1) = 0 and f''(1) < 0; and a relative minimum at x = 3 because f'(3) = 0 and f''(3) > 0.
Example 3 — The inconclusive case (NO calculator)
Problem. Classify the critical point(s) of f(x) = x⁴. Show why the Second Derivative Test fails and resolve it.
Strategy. Apply the test; when it returns 0, switch to the First Derivative Test.
Solution.
f'(x) = 4x³ ⇒ f'(x) = 0 at x = 0 (only critical point)
f''(x) = 12x² ⇒ f''(0) = 0 → INCONCLUSIVEThe Second Derivative Test gives no information. Return to the First Derivative Test — check the sign of f' on each side of 0:
x: −1 1
f'(x): 4(−1)³ 4(1)³
= −4 = +4
Sign: − +f' changes from negative to positive at x = 0.
Justification. By the First Derivative Test, f has a relative minimum at x = 0 because f' changes from negative to positive there. (The Second Derivative Test was inconclusive because f''(0) = 0.)
Example 4 — Deduce concavity from the graph of f' (NO calculator)
Problem. The graph of f' (the derivative, not f) is given below. On what interval is f concave down, and where does f have an inflection point? Justify.
![graph of f'(x) on [0, 6] × [−3, 3] — this is f-PRIME](calciq-graphs/L22-1.svg)
Strategy. Concavity of f is governed by whether f' is increasing or decreasing (that is the sign of f''). Inflection points of f occur where f' turns around (local extremum of f').
Solution. From the graph of f':
f'is decreasing on(0, 2)⇒f''<0⇒fis concave DOWN on(0, 2).f'is increasing on(2, 6)⇒f''>0⇒fis concave UP on(2, 6).
Justification. f has an inflection point at x = 2 because f' changes from decreasing to increasing there, so f'' changes from negative to positive (concavity changes from down to up). Note the crossing at x = 4 is where f has a relative minimum (f' changes − → +), not an inflection point — a classic mix-up. The inflection comes from the turning point of f' (at x = 2), not its zero.
(d) Common Mistakes
Treating f''(x) = 0 as an automatic inflection point. What students do: solve f'' = 0, declare every solution an inflection point. Why it's wrong: f'' = 0 is only a candidate; if f'' doesn't change sign (e.g., f''=12x² at x=0), there is no inflection. Fix: always test the sign of f'' on both sides and confirm it changes.
Confusing concavity with increasing/decreasing. What students do: say "f is concave up because f is increasing." Why it's wrong: increasing/decreasing is the sign of f'; concavity is the sign of f''. A function can increase while concave down (e.g. √x). Fix: concavity = behavior of f', one derivative deeper.
Flipping the Second Derivative Test signs. What students do: think f''>0 means maximum. Why it's wrong: f''>0 is a cup ∪, so a horizontal tangent there is a minimum. Fix: anchor it — "positive second derivative = concave up = bottom of a cup = MIN."
Using f' where f'' is needed (or vice versa). What students do: find inflection points from f'=0, or extrema from f''=0. Why it's wrong: extrema come from f'; concavity/inflection come from f''. Fix: f' answers "increasing/decreasing & extrema"; f'' answers "concavity & inflection."
Reading a graph of f' as if it were f. What students do: on an FRQ giving the graph of f', call its x-intercepts inflection points. Why it's wrong: on the graph of f', the zeros mark candidate extrema of f; the turning points (local max/min) of f' mark inflection points of f. Fix: slow down and label which function you are looking at.
(e) Practice Problems
f(x) = x³ − 3x², on which interval is the graph concave up?f''(x) = 6x − 12, the graph of f has an inflection point at x =f occurs at x = 4, where f'(4) = 0 and f''(4) = −5. The Second Derivative Test shows f has ac, f'(c) = 0 and f''(c) = 0. Which is the correct next step?f(x) = x⁴ − 6x², how many inflection points does the graph have?f' is decreasing on an interval, then on that interval f isf(x) = x⁵ − 5x³, use your calculator to find the largest x-coordinate of an inflection point.f(x) = e^{−x²}. Find the positive x-value where f has an inflection point (round to three decimals).Justify. For f(x) = x³ − 12x, find all critical points and classify each using the Second Derivative Test. Show the value of f'' at each critical point and write a one-sentence AP-style justification.
Justify. Explain in one or two sentences why f''(c) = 0 does not guarantee that f has an inflection point at x = c. Give a specific function as a counterexample.
Graph interpretation. The graph of f' rises to a local maximum at x = −1, falls to a local minimum at x = 3, and rises afterward. State the interval(s) on which f is concave up and identify the x-coordinate(s) of every inflection point of f. Justify.
For f(x) = x⁴ − 4x³ + 10, find the interval(s) of concavity and all inflection points. Show your f'' sign chart.
Let f(x) = x + 2sin x on [0, 2π]. Use your calculator to find the x-coordinates of all inflection points on this interval.
(f) AP Exam Focus
> Free-Response Question (No calculator). 9 points total.
>
> Let f be a twice-differentiable function defined for all real numbers. The graph of f', the derivative of f, is shown below on the interval [−2, 5].
>
> `
> ![graph of f'(x) on [−2, 5] × [−4, 4] — this is f-PRIME, the derivative >](calciq-graphs/L22-2.svg)
> `
>
> (a) On what open interval(s) is f increasing? Justify your answer. (2 points)
>
> (b) Find the x-coordinate of each critical point of f and use the First Derivative Test to classify it as a relative maximum, relative minimum, or neither. Justify your answer. (2 points)
>
> (c) On what open interval(s) is the graph of f concave up? Give the x-coordinate of every inflection point of f, and justify. (3 points)
>
> (d) At x = 1, is f increasing or decreasing, and is the graph of f concave up or concave down? Justify each answer. (2 points)
Model Solution
(a) f is increasing where f'(x) > 0. From the graph, f' is positive on (−2, 0) and on (2, 5).
f is increasing on (−2, 0) and (2, 5) because f'(x) > 0 on those intervals.
(b) Critical points occur where f'(x) = 0: at x = 0 and x = 2.
- At x = 0, f' changes from positive to negative ⇒ relative maximum.
- At x = 2, f' changes from negative to positive ⇒ relative minimum.
By the First Derivative Test, f has a relative maximum at x = 0 because f' changes from positive to negative there, and a relative minimum at x = 2 because f' changes from negative to positive there.
(c) The graph of f is concave up where f''(x) > 0, i.e., where f' is increasing. From the graph, f' is increasing on (1, 4).
The graph of f is concave up on (1, 4) because f' is increasing there (so f'' > 0).
Inflection points occur where f'' changes sign — i.e., where f' changes between increasing and decreasing (the turning points of f'):
- At x = 1, f' changes from decreasing to increasing ⇒ f'' changes from − to +.
- At x = 4, f' changes from increasing to decreasing ⇒ f'' changes from + to −.
f has inflection points at x = 1 and x = 4 because f'' changes sign at each (f' changes from decreasing to increasing at x = 1, and from increasing to decreasing at x = 4).
(d) At x = 1: the graph shows f'(1) = −3 < 0, so f is decreasing at x = 1. Also, x = 1 is the local minimum of f', and just to its right f' is increasing, so f'' (1) = 0 and f'' is transitioning — but the question asks about the immediate behavior: at x = 1, f' is at its minimum (a turning point), the inflection point.
At x = 1, f is decreasing because f'(1) = −3 < 0. The point x = 1 is an inflection point of f (the local minimum of f'), where concavity changes from down to up; it is the boundary, so f is neither concave up nor concave down exactly at x = 1 (concave down just to the left, concave up just to the right).
Scoring Commentary
- (a) [2 pts]: 1 pt for both correct intervals (−2,0) and (2,5); 1 pt for the justification "f' > 0." Students lose the justification point for just stating intervals with no reason.
- (b) [2 pts]: 1 pt for both critical points with correct classification; 1 pt for First-Derivative-Test justification citing the sign change of f'. Common loss: saying "max because the graph goes down" without referencing f''s sign change.
- (c) [3 pts]: 1 pt for concave-up interval (1,4); 1 pt for both inflection x-values 1 and 4; 1 pt for justification "f' changes increasing/decreasing ⇒ f'' changes sign." Biggest trap: students name the zeros of f' (x=0, 2) as inflection points — those are extrema of f, worth 0 for this part.
- (d) [2 pts]: 1 pt for "decreasing, because f'(1) < 0"; 1 pt for correctly addressing concavity at the turning point of f'. Accept "inflection point / concavity changing" with justification.
🔑 Answer Key
1. (D). f''(x) = 6x − 6; f'' > 0 when x > 1. (B) is concave-down; (A)/(C) confuse concavity with the increasing/decreasing or domain split.
2. (A). f'' = 6x − 12 = 0 at x = 2, and f'' changes − → + there. (C) confuses this with a f' = 0 style answer; (B) is arbitrary; (D) is wrong since the sign does change.
3. (A). f'(4) = 0 and f''(4) < 0 ⇒ concave down ⇒ relative maximum. (D) flips the sign rule; (C) confuses the test's purpose; (B) is wrong because f''(4) ≠ 0.
4. (B). f''(c) = 0 makes the Second Derivative Test inconclusive, so apply the First Derivative Test. (D) wrongly assumes inflection; (C)/(A) jump to a conclusion the data don't support.
5. (C). f(x) = x⁴ − 6x² ⇒ f''(x) = 12x² − 12 = 12(x²−1) = 12(x−1)(x+1), zero at x = ±1, sign changing at both ⇒ 2 inflection points. (A) ignores them; (B) miscounts; (D) over-counts.
6. (D). f' decreasing ⇒ f'' < 0 ⇒ concave down. (C) flips it; (A)/(B) confuse concavity with the sign of f' itself.
7. (C). f(x) = x⁵ − 5x³ ⇒ f''(x) = 20x³ − 30x = 10x(2x² − 3), with zeros at x = 0 and 2x² − 3 = 0 ⇒ x² = 1.5 ⇒ x = ±√1.5 ≈ ±1.225. All three values are inflection points (each is a sign change); the largest is √1.5 ≈ 1.225. TI-84: graph f'' and find the rightmost zero. (A) is the middle inflection, not the largest; (B) √3 and (D) √6 come from mis-solving 2x² − 3 = 0 (forgetting to divide by 2, or using x² = 6).
8. (B). f(x) = e^{−x²} ⇒ f'(x) = −2x e^{−x²} ⇒ f''(x) = e^{−x²}(4x² − 2). f'' = 0 when 4x² − 2 = 0 ⇒ x² = 1/2 ⇒ x = 1/√2 ≈ 0.707. TI-84: solve nDeriv(nDeriv(e^(−X²),X,X),X) = 0, or graph f'' and find the positive zero. (A) misreads x²=1/4; (C)/(D) come from x²=1 or x²=2.
9. f(x) = x³ − 12x ⇒ f'(x) = 3x² − 12 = 3(x²−4), zero at x = −2 and x = 2. f''(x) = 6x.
- f''(−2) = −12 < 0 ⇒ relative maximum at x = −2.
- f''(2) = 12 > 0 ⇒ relative minimum at x = 2.
By the Second Derivative Test, f has a relative maximum at x = −2 because f'(−2) = 0 and f''(−2) < 0, and a relative minimum at x = 2 because f'(2) = 0 and f''(2) > 0.
10. f''(c) = 0 is only a necessary candidate condition — an inflection point also requires f'' to change sign at c. Counterexample: f(x) = x⁴ has f''(x) = 12x², so f''(0) = 0, but f'' is positive on both sides of 0 (concave up throughout), so there is no inflection point at x = 0.
11. The graph of f' rises to a local max at x = −1, falls to a local min at x = 3, then rises. So f' is increasing on (−∞, −1) and (3, ∞) and decreasing on (−1, 3).
- f is concave up on (−∞, −1) and (3, ∞) because f' is increasing (f'' > 0) there.
- Inflection points at x = −1 and x = 3, because f' turns around at each (f'' changes sign): + → − at x = −1 and − → + at x = 3.
12. f(x) = x⁴ − 4x³ + 10 ⇒ f'(x) = 4x³ − 12x² ⇒ f''(x) = 12x² − 24x = 12x(x − 2). Zeros at x = 0, 2.
Interval: (−∞, 0) (0, 2) (2, ∞)
f'' sign: + − +
Concavity: UP DOWN UPConcave up on (−∞, 0) and (2, ∞); concave down on (0, 2). Inflection points at x = 0 (point (0, 10)) and x = 2 (point (2, 16 − 32 + 10) = (2, −6)), since f'' changes sign at each.
13. f(x) = x + 2sin x ⇒ f'(x) = 1 + 2cos x ⇒ f''(x) = −2 sin x. On [0, 2π], f'' = 0 where sin x = 0: x = 0, π, 2π. Sign change of −2 sin x occurs at the interior point x = π (where sin x crosses zero). So the only inflection point on the open interior is x = π ≈ 3.142. (TI-84: graph f''(x) = −2sin(X) and find its zero with a sign change; the endpoints 0 and 2π are not interior inflection points.)
CalcIQ · Lesson 22 of 35 · Unit 5 — Analytical Applications of Differentiation. Built and reviewed by a mathematician.
Disclaimer: AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this product. This lesson is independent study material for the AP Calculus AB exam.
Accuracy review note: All derivatives, f'' sign charts, inflection points, and Second Derivative Test classifications in this lesson were independently recomputed and verified. One judgment call flagged for Isaac's QC: FRQ part (d) phrasing — at x = 1 the function is decreasing (f'(1) < 0) and x = 1 is itself the inflection point, so f is neither concave up nor down exactly there; the model answer states this explicitly rather than forcing a single concavity label.