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Lesson 66 — Concavity and Inflection Points

Sign of f'': concave up when f'' > 0, down when f'' < 0. Inflection where f'' changes sign. Second derivative test for extrema.

Used in: 2nd Year High School Advanced · Japanese Equiv. Math I/II · German Equiv. Leistungskurs Analysis · University Calculus I

f''(x) > 0 \implies f \text{ concave}\uparrow, \quad f''(x) < 0 \implies f \text{ concave}\downarrow, \quad f'' \text{ changes sign} \implies \text{inflection}

The concavity of a curve is determined by the sign of the second derivative: $f'' > 0$ means concave up (bowl shape), $f'' < 0$ concave down (cap shape). An inflection point occurs where $f''$ changes sign — it is not enough for it to be zero.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous Definition and Criteria

Concavity and Convexity

"The function $f$ is concave up on an interval $I$ if $f''(x) \geq 0$ for all $x \in I$ ." — OpenStax Calculus Vol. 1 §4.5

Criterion via Second Derivative: If $f$ is twice differentiable on $I$ :

$f''(x) \geq 0$ on $I$ $\iff$ $f$ is convex (concave up).
$f''(x) \leq 0$ on $I$ $\iff$ $f$ is concave (downward).
$f''(x) > 0$ strictly $\Rightarrow$ strict convexity.

Concave up (f'' > 0): chord lies above the arc. Concave down (f'' < 0): chord lies below the arc.

Inflection Point

Caution: $f''(x_0) = 0$ is a necessary but NOT sufficient condition. Canonical counterexample: $f(x) = x^4$ has $f''(0) = 0$ but $f'' \geq 0$ in a neighborhood of $0$ — no sign change, thus $0$ is not an inflection point.

"If the concavity changes at a point $(x_0, f(x_0))$ , we call this a point of inflection. It must be the case that $f''(x_0)$ changes sign." — APEX Calculus §3.4

Second Derivative Test for Local Extrema

Proof for minimum: If $f'(x_0) = 0$ and $f''(x_0) > 0$ , by continuity of $f''$ there exists a neighborhood where $f''(x) > 0$ , hence $f'$ is increasing in this neighborhood. Since $f'(x_0) = 0$ , we have $f' < 0$ to the left and $f' > 0$ to the right of $x_0$ — by the first derivative test, $x_0$ is a local minimum. ∎

Solved Examples

Example— 1· Concavity and Inflection of f(x) = x^3 - 3x (Basic)

Problem. For $f(x) = x^3 - 3x$ , determine the intervals of concavity and the inflection points.

Strategy. Calculate $f''$ , determine where it is positive/negative, and check for sign changes at zeros.

Solution.

Derivatives: $f'(x) = 3x^2 - 3$ , $f''(x) = 6x$ .
Zero of $f''$ : $6x = 0 \Rightarrow x = 0$ .
Sign: $f''(x) < 0$ for $x < 0$ (concave down); $f''(x) > 0$ for $x > 0$ (concave up).
Inflection: $f''$ changes sign at $x = 0$ . Thus $(0, f(0)) = (0, 0)$ is an inflection point.

Verification. In the sketch: $f(-2) = -8 + 6 = -2$ ; $f(0) = 0$ ; $f(2) = 8 - 6 = 2$ . The curve rises from $-2$ to $0$ concave down, then continues from $0$ to $2$ concave up — consistent. ✓

Source. APEX Calculus §3.4, Example 3.4.1 — CC-BY-NC.

Example— 2· Second Derivative Test for Extrema of f(x) = x^4 - 4x^2 (Intermediate)

Problem. Classify the critical points of $f(x) = x^4 - 4x^2$ using the second derivative test.

Strategy. Find critical points via $f' = 0$ , calculate $f''$ at each, and apply the criterion.

Solution.

$f'(x) = 4x^3 - 8x = 4x(x^2 - 2) = 0 \Rightarrow x = 0,\, x = \pm\sqrt{2}$ .
$f''(x) = 12x^2 - 8$ .
Evaluating:
- $f''(0) = -8 < 0$ $\Rightarrow$ local maximum at $x = 0$ . $f(0) = 0$ .
- $f''(\sqrt{2}) = 24 - 8 = 16 > 0$ $\Rightarrow$ local minimum at $x = \sqrt{2}$ . $f(\sqrt{2}) = 4 - 8 = -4$ .
- $f''(-\sqrt{2}) = 16 > 0$ $\Rightarrow$ local minimum at $x = -\sqrt{2}$ . $f(-\sqrt{2}) = -4$ .

Verification. As $f(x) \to +\infty$ for $x \to \pm\infty$ , the minima at $\pm\sqrt{2}$ are global, and the maximum at $0$ is local (but not global). ✓

Source. OpenStax Calculus Vol. 1 §4.5, Example 4.38 — CC-BY-NC-SA.

Example— 3· Inflection Points of the Gaussian (Intermediate)

Problem. Find the inflection points of $f(x) = e^{-x^2/2}$ (the kernel of the normal distribution).

Strategy. Differentiate twice, set $f'' = 0$ , and check for sign changes.

Solution.

$f'(x) = -x\,e^{-x^2/2}$ .
$f''(x) = -e^{-x^2/2} + x^2 e^{-x^2/2} = (x^2 - 1)e^{-x^2/2}$ .
Zero of $f''$ : $x^2 - 1 = 0 \Rightarrow x = \pm 1$ .
Sign:
- $|x| > 1$ : $x^2 - 1 > 0 \Rightarrow f'' > 0$ (concave up).
- $|x| < 1$ : $x^2 - 1 < 0 \Rightarrow f'' < 0$ (concave down).
Inflection points: at $x = \pm 1$ , $f''$ changes sign. Points: $(\pm 1, e^{-1/2}) = (\pm 1, \approx 0.607)$ .

Verification. For $\mathcal{N}(\mu, \sigma^2)$ , inflection points are at $\mu \pm \sigma$ . Here $\mu = 0$ , $\sigma = 1$ — consistent. ✓

Source. Active Calculus §3.1, Activity 3.1.3 — CC-BY-NC-SA.

Example— 4· Inconclusive Case: f'' = 0 at Critical Point is Not Inflection (Advanced)

Problem. For $f(x) = x^4$ , verify that $f''(0) = 0$ but $x = 0$ is a local minimum (not an inflection point).

Strategy. Show that $f''$ does not change sign at $x = 0$ .

Solution.

$f'(x) = 4x^3 = 0 \Rightarrow x = 0$ (critical point).
$f''(x) = 12x^2 \geq 0$ for all $x$ . In particular $f''(0) = 0$ — test is inconclusive.
But $f''(x) \geq 0$ in a neighborhood of $0$ : no sign change.
Conclusion: $x = 0$ is NOT an inflection point. It is a global minimum ( $f(x) = x^4 \geq 0$ for all $x$ , with equality only at $0$ ).
Confirmation by $f'$ sign test: $f'(x) = 4x^3 < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$ — minimum confirmed.

Verification. The graph of $x^4$ is a parabola "flattened" at the bottom at $0$ — visually a minimum. ✓

Source. APEX Calculus §3.4, Note after Example 3.4.2 — CC-BY-NC.

Example— 5· Decreasing Returns Cost — Economic Inflection (Modeling)

Problem. A company has production cost $C(q) = q^3 - 6q^2 + 15q + 100$ (in dollars). Find the inflection point and interpret it economically.

Strategy. Calculate $C''$ , find the inflection point, and analyze the behavior of the marginal cost $MC = C'$ .

Solution.

$C'(q) = 3q^2 - 12q + 15 = MC(q)$ (marginal cost).
$C''(q) = 6q - 12$ .
Inflection: $C''(q) = 0 \Rightarrow q = 2$ .
Sign of $C''$ : for $q < 2$ , $C'' < 0$ (decreasing marginal cost — economies of scale). For $q > 2$ , $C'' > 0$ (increasing marginal cost — capacity pressure).
Inflection point: $C(2) = 8 - 24 + 30 + 100 = 114$ . Point $(2, 114)$ .

Verification. $MC(0) = 15$ , $MC(2) = 12 - 24 + 15 = 3$ , $MC(4) = 48 - 48 + 15 = 15$ . Minimum $MC$ at $q = 2$ — consistent with inflection of $C$ . ✓

Source. Active Calculus §3.1, Exercise 3.1.5 — CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

20 10 4 2 4

Ex. 66.1
Determine the concavity of $f(x) = x^2$ on all of $\mathbb{R}$ . Is there an inflection point?
Solve online APEX Calculus · §3.4
Ex. 66.2
Determine the concavity and inflection points of $f(x) = x^3$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
$f'' = 6x$ . For $x < 0$ : concave down. For $x > 0$ : concave up. Inflection at $x = 0$ because $f''$ changes sign.
Show step-by-step (with the why)
1. Calculate derivatives: $f'(x) = 3x^2$ , $f''(x) = 6x$ .
2. Zero of $f''$ : $6x = 0 \Rightarrow x = 0$ .
3. Sign: $x < 0 \Rightarrow f'' < 0$ (cap); $x > 0 \Rightarrow f'' > 0$ (bowl).
4. Sign change at $x = 0$ confirms inflection. Point: $(0, 0)$ .
5. Curiosity: the inflection point of $x^3$ is also its only zero — the curve crosses the x-axis exactly where it changes concavity.
Ex. 66.3
Concavity of $f(x) = x^4$ . Is there an inflection point at $x = 0$ ? Justify using the sign of $f''$ .
Solve online APEX Calculus · §3.4
Show solution
$f'' = 12x^2 \geq 0$ for all $x$ . No sign change — no inflection. Concave up on all of $\mathbb{R}$ (global convexity).
Ex. 66.4
Concavity of $f(x) = e^x$ on all of $\mathbb{R}$ . Is there an inflection point?
Solve online OpenStax Calculus Vol. 1 · §4.5
Ex. 66.5Answer key
Concavity of $f(x) = \ln x$ on $(0, \infty)$ .
Solve online APEX Calculus · §3.4
Ex. 66.6
Concavity of $f(x) = \sin x$ on $[0, 2\pi]$ . Identify the inflection points.
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
$f'' = -\sin x$ . Zero at $x = 0, \pi, 2\pi$ . Inflection points at $x = 0, \pi, 2\pi$ (where $f''$ changes sign). For $x \in (0, \pi)$ : $f'' < 0$ (cap). For $x \in (\pi, 2\pi)$ : $f'' > 0$ (bowl).
Ex. 66.7
Concavity of $f(x) = \cos x$ on $[0, 2\pi]$ . Inflection points.
Solve online APEX Calculus · §3.4
Ex. 66.8
Concavity of $f(x) = 1/x$ on the intervals $(0,\infty)$ and $(-\infty,0)$ .
Solve online Active Calculus · §3.1
Ex. 66.9
Concavity of $f(x) = e^{-x^2/2}$ (Gaussian). Identify the inflection points.
Solve online OpenStax Calculus Vol. 1 · §4.5
Ex. 66.10
Concavity and inflection of $f(x) = x^3 - 3x^2 + 2$ .
Solve online APEX Calculus · §3.4
Show solution
$f'' = 6x - 6$ . Zero at $x = 1$ . Inflection at $(1, f(1)) = (1, 1 - 3 + 2) = (1, 0)$ . Concave down on $(-\infty, 1)$ , up on $(1, \infty)$ .
Show step-by-step (with the why)
1. Calculate $f'(x) = 3x^2 - 6x$ and $f''(x) = 6x - 6$ .
2. Zero of $f''$ : $6x - 6 = 0 \Rightarrow x = 1$ .
3. Sign: $x < 1 \Rightarrow f'' < 0$ ; $x > 1 \Rightarrow f'' > 0$ . Change confirmed.
4. Point: $f(1) = 1 - 3 + 2 = 0$ . Inflection at $(1, 0)$ .
5. Shortcut: for cubic polynomials $ax^3 + bx^2 + \cdots$ , the inflection point is always at $x = -b/(3a)$ — the center of symmetry of the cubic.
Ex. 66.11Answer key
Use the $f''$ test: classify the extrema of $f(x) = x^3 - 12x$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
$f' = 3x^2 - 12 = 0 \Rightarrow x = \pm 2$ . $f'' = 6x$ . $f''(2) = 12 > 0$ : minimum at $(2, -16)$ . $f''(-2) = -12 < 0$ : maximum at $(-2, 16)$ .
Ex. 66.12
Extrema of $f(x) = x^4 - 4x^2$ via $f''$ test.
Solve online APEX Calculus · §3.4
Ex. 66.13
Extrema of $f(x) = x e^{-x}$ via $f''$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
$f'(x) = e^{-x}(1-x) = 0 \Rightarrow x = 1$ . $f''(x) = e^{-x}(x-2)$ . $f''(1) = e^{-1}(-1) < 0$ : maximum at $(1, e^{-1})$ .
Ex. 66.14Answer key
Extrema of $f(x) = x^2 \ln x$ on $(0, \infty)$ via $f''$ .
Solve online Active Calculus · §3.1
Ex. 66.15Answer key
Extrema of $f(x) = \sin x + \frac{1}{2}\sin(2x)$ on $[0, 2\pi]$ .
Solve online APEX Calculus · §3.4
Show solution
$f'(x) = \cos x + \cos(2x) = 0$ . Using $\cos(2x) = 2\cos^2 x - 1$ : $2\cos^2 x + \cos x - 1 = 0 \Rightarrow \cos x = 1/2$ or $-1$ . Critical points: $x = \pi/3, \pi, 5\pi/3$ . Apply $f''$ at each.
Ex. 66.16
Show that $f(x) = x^4$ has a minimum at $x = 0$ despite $f''(0) = 0$ (inconclusive test).
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
Show that $x = 0$ is a minimum of $x^4$ even with $f''(0) = 0$ : use the fact that $f(x) = x^4 \geq 0$ for all $x$ , and $f(0) = 0$ is the smallest possible value.
Ex. 66.17
Show that $f(x) = x^5$ has no extremum at $x = 0$ despite $f'(0) = 0$ .
Solve online APEX Calculus · §3.4
Show solution
For $f(x) = x^5$ : $f'(0) = 0$ and $f''(0) = 0$ (inconclusive). But $f'(x) = 5x^4 \geq 0$ for all $x$ , with equality only at $0$ . Therefore $f$ is increasing in any neighborhood of $0$ — no local extremum.
Ex. 66.18
For $f(x) = x^2 + 1/x$ on $x > 0$ : find the minimum and justify with $f''$ .
Solve online Active Calculus · §3.1
Ex. 66.19
Extrema of $f(x) = \ln x / x$ on $(0, \infty)$ via $f''$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
$f'(x) = (1 - \ln x)/x^2 = 0 \Rightarrow x = e$ . $f''(e) = (-3 + 2)/e^3 = -1/e^3 < 0$ : maximum. $f(e) = 1/e$ .
Ex. 66.20
Extrema of $f(x) = x^{1/x}$ on $(0, \infty)$ (take $\ln f$ before differentiating).
Solve online APEX Calculus · §3.4
Ex. 66.21
Cost $C(q) = q^3 - 6q^2 + 9q + 100$ . Find the inflection point and interpret it as a change in marginal return.
Solve online Active Calculus · §3.1
Show solution
$C''(q) = 6q - 12 = 0 \Rightarrow q = 2$ . For $q < 2$ : decreasing marginal cost (economies of scale). For $q > 2$ : increasing (capacity pressure). Inflection at $q = 2$ is the point of regime change.
Ex. 66.22Answer key
Profit $\pi(q) = -q^3 + 30q^2 - 100q$ . Maximize via $\pi'$ and confirm with $\pi''$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Ex. 66.23Answer key
Logistic curve $P(t) = K/(1 + e^{-rt})$ . Show there is an inflection point at $P = K/2$ (half the carrying capacity).
Solve online Active Calculus · §3.1
Show solution
For $P(t) = K/(1 + e^{-rt})$ : $P'' = rP(r - 2rP/K)$ . Set to zero: $P = K/2$ , which occurs at $t = 0$ (when centered at zero). Inflection at $P = K/2$ : point of maximum population growth.
Show step-by-step (with the why)
1. Let $P = K/(1+e^{-rt})$ . Calculate $P' = rP(1 - P/K)$ (logistic growth rate).
2. Differentiate again: $P'' = r P' (1 - 2P/K)$ .
3. Set to zero: $P = 0$ (trivial) or $P = K/2$ .
4. At $P = K/2$ : $P''$ changes sign — inflection.
5. Note: The inflection point of the logistic curve corresponds to the maximum growth rate — the "turning point" of an epidemic or market expansion.
Ex. 66.24
Potential energy $U(x) = -\cos x$ (pendulum). Find stable and unstable equilibria using $U''$ .
Solve online APEX Calculus · §3.4
Ex. 66.25
Harmonic spring: $U(x) = \frac{1}{2}kx^2$ . Show $x = 0$ is a stable equilibrium using $U''$ .
Solve online Active Calculus · §3.1
Show solution
$U(x) = \frac{1}{2}kx^2$ . $U'(x) = kx = 0$ at $x = 0$ . $U''(x) = k > 0$ : concave up. Minimum at $x = 0$ — stable equilibrium. Natural oscillation frequency: $\omega = \sqrt{k/m}$ .
Ex. 66.26
Bernoulli entropy $H(p) = -p\ln p - (1-p)\ln(1-p)$ . Show $H'' < 0$ and that the maximum is at $p = 1/2$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Ex. 66.27Answer key
Learning curve $L(t) = 1 - e^{-kt}$ . Determine its concavity. What does it say about the speed of learning?
Solve online Active Calculus · §3.1
Show solution
Learning curve: $L(t) = 1 - e^{-kt}$ . $L'' = -k^2 e^{-kt} < 0$ . Concave down: learning gains diminish over time (diminishing returns). No inflection.
Ex. 66.28
In an epidemic, the peak of new cases occurs at the inflection point of the cumulative case curve $f(t)$ . Justify geometrically and via $f''$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Ex. 66.29
Utility $U(W) = \ln W$ is concave. Explain how Jensen's inequality implies risk aversion for this investor.
Solve online Active Calculus · §3.1
Show solution
Utility $U(W) = \ln W$: $U'' = -1/W^2 < 0$ — concave down. By Jensen: $E[U(W)] \leq U(E[W])$ . This means the agent prefers the guaranteed expected value to the lottery — definition of risk aversion.
Ex. 66.30
Why does the linear regression loss function have a unique global minimum? Justify using convexity.
Solve online Active Calculus · §3.1
Ex. 66.31
What is the correct condition for $x_0$ to be an inflection point of $f$ ?
Select the correct option
$f''(x_0) = 0$ and $f''$ changes sign passing through $x_0$ . $f''(x_0) = 0$ and $f'(x_0) = 0$ . $f''(x_0) = 0$ , regardless of the sign around it. $f'(x_0) = 0$ and $f''(x_0) = 0$ , but $f'''(x_0) eq 0$ .
Select an option first
Solve online APEX Calculus · §3.4
Show solution
The definition of an inflection point requires a sign change of $f''$ . Having $f'' = 0$ without a sign change (like for $x^4$ at 0) does not constitute an inflection.
Ex. 66.32
Prove that the sum of two convex functions is convex, using the definition via $f''$ .
Active Calculus · §3.1
Show solution
Sum: $(f+g)'' = f'' + g''$ . If $f'' \geq 0$ and $g'' \geq 0$ , then $(f+g)'' \geq 0$ . Thus $f+g$ is convex.
Ex. 66.33
Show that $f$ convex on $I$ implies the midpoint inequality: $f\!\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$ .
Solve online OpenStax Calculus Vol. 1 · §4.5
Show solution
Show that $f((x+y)/2) \leq (f(x)+f(y))/2$ : use the definition of convexity with $t = 1/2$ . Geometrically: the midpoint of the segment lies above the midpoint of the graph.
Ex. 66.34
Why is $f''(x_0) = 0$ not sufficient to guarantee an inflection point? Give a concrete counterexample.
Solve online APEX Calculus · §3.4
Show solution
Provide the counterexample $f(x) = x^4$: $f''(0) = 0$ but $f'' \geq 0$ around $0$ — no sign change. Thus $x = 0$ is not an inflection point despite $f'' = 0$ .
Ex. 66.35
Show that $\ln$ is concave on $(0,\infty)$ and use it to prove the AM-GM inequality: $(x+y)/2 \geq \sqrt{xy}$ for $x, y > 0$ .
Solve online Active Calculus · §3.1
Show solution
$\ln$ is concave: $\ln((x+y)/2) \geq (\ln x + \ln y)/2 = \ln\sqrt{xy}$ . Applying the exponential: $(x+y)/2 \geq \sqrt{xy}$ . QED.
Ex. 66.36Answer key
Huber function $L(x) = x^2/2$ if $|x| \leq 1$ ; $|x| - 1/2$ otherwise. Is it convex? Where is $L''$ discontinuous?
Solve online APEX Calculus · §3.4
Show solution
Huber loss: $L(x) = x^2/2$ if $|x| \leq 1$ , $|x| - 1/2$ otherwise. Second derivative: $L'' = 1$ for $|x| < 1$ , $L'' = 0$ for $|x| > 1$ . $L''$ does not exist at $x = \pm 1$ . It is convex (non-negative). It is not strictly convex.
Ex. 66.37
Prove the second derivative test using the Taylor polynomial of order 2.
OpenStax Calculus Vol. 1 · §4.5
Show solution
Taylor expansion of order 2 around $x_0$ : $f(x) \approx f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2}(x-x_0)^2$ . With $f'(x_0) = 0$ : $f(x) - f(x_0) \approx \frac{f''(x_0)}{2}(x-x_0)^2$ . If $f''(x_0) > 0$ : the right side is positive near $x_0$ , so $f(x) > f(x_0)$ — local minimum. Similarly for $f'' < 0$ .
Ex. 66.38Answer key
Prove Jensen's inequality for two points: $f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$ — directly from the definition of convexity.
Active Calculus · §3.1
Show solution
Definition of convexity: $f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$ for all $t \in [0,1]$ . With $t = 1/2$ and properties of integrals/probability, the general Jensen inequality can be deduced by induction or continuity.
Ex. 66.39
Prove that a convex function on an open interval is continuous on the interior.
APEX Calculus · §3.4
Show solution
Let $f$ be convex on an open interval $I$ . For $x_0 \in I$ , the quotient function $(f(x) - f(x_0))/(x - x_0)$ is increasing in $x$ . This implies that the one-sided limits exist and are finite, hence $f$ is continuous at $x_0$ .
Ex. 66.40Answer key
Prove that $f$ is convex if and only if its graph lies above every tangent line: $f(y) \geq f(x) + f'(x)(y-x)$ for all $x, y$ .
Active Calculus · §3.1
Show solution
From the Taylor remainder of order 1 with expansion point at $x$ : $f(y) \geq f(x) + f'(x)(y - x)$ for all $y$ , since $R_1(y) = f''(\xi)(y-x)^2/2 \geq 0$ . This is equivalent to convexity (support by the tangent).

Sources

Active Calculus — Boelkins · 2024 · §3.1 Using Derivatives to Identify Extreme Values · CC-BY-NC-SA. Primary source.
Calculus Volume 1 — OpenStax · 2016 · §4.5 Derivatives and the Shape of a Graph · CC-BY-NC-SA.
APEX Calculus — Hartman et al. · 2024 · v5.0 · §3.4 Concavity and the Second Derivative Test · CC-BY-NC.