Worked examples — Second derivative test — concavity, inflection points

This page is a drill hall. The parent note built the tools; here we fire every kind of question at them. Before we solve anything, we lay out a matrix of every case class the topic can throw at you — then each worked example is tagged with the exact cell it covers, so by the end there is no scenario you have not seen.

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The scenario matrix

Every problem in this topic falls into one of these cells. The worked examples afterward are labelled by cell so you can see the coverage is complete.

Cell	Case class	What makes it tricky	Example
C1	$f^{\prime \prime }(c)>0$ at a critical point	clean minimum	(A)
C2	$f^{\prime \prime }(c)<0$ at a critical point	clean maximum	(A)
C3	$f^{\prime \prime }(c)=0$, sign does flip	genuine inflection	(A), (C)
C4	$f^{\prime \prime }(c)=0$, sign does not flip	trap: no inflection, test fails	(B)
C5	$f^{\prime \prime }$ does not exist at $c$	inflection with a corner/cusp in $f^{\prime \prime }$	(D)
C6	Multiple/periodic inflections	trig, infinitely many switches	(C)
C7	Endpoint / restricted domain	2nd-deriv test says nothing at edges	(E)
C8	Real-world optimization word problem	translate words → $f$, then classify	(F)
C9	Exam twist: unknown parameter	choose a constant so a point is an inflection	(G)

We will hit all nine cells across seven examples.

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Example (A) — Cells C1, C2, C3: the full sweep on a cubic

Step 1 — find where the road goes flat. $f^{\prime }(x)=3x^2-3=0\Rightarrow x^2=1\Rightarrow x=\pm 1.$ Why this step? Critical points ($f^{\prime }=0$) are the only places a local max or min can hide — the tangent is horizontal there.

Step 2 — get the concavity tool. $f^{\prime \prime }(x)=6x.$ Why this step? At a flat point, the sign of $f^{\prime \prime }$ decides cup vs cap, hence min vs max — no need to test neighbours.

Step 3 — apply the test at each critical point (Cells C1, C2). $f^{\prime \prime }(1)=6>0\Rightarrow \cup \Rightarrow \text{local min at }x=1,f(1)=1-3=-2.$ $f^{\prime \prime }(-1)=-6<0\Rightarrow \cap \Rightarrow \text{local max at }x=-1,f(-1)=-1+3=2.$ Why this step? "Up is a cuP, plus on top" — positive $f^{\prime \prime }$ is a valley bottom, negative is a hill top.

Step 4 — inflection candidate (Cell C3). $f^{\prime \prime }(x)=6x=0\Rightarrow x=0.$ Sign check: for $x<0$, $f^{\prime \prime }<0$ ($\cap$); for $x>0$, $f^{\prime \prime }>0$ ($\cup$). It flips, so $(0,f(0))=(0,0)$ is a genuine inflection. Why this step? $f^{\prime \prime }=0$ is only a candidate; concavity must actually switch.

Figure s01 — read the whole cubic at a glance. The plot below shows all three features on one curve: the orange dot at $(0,0)$ is the inflection where the road switches from frown ($\cap$) to smile ($\cup$); the red dot at $(-1,2)$ sits at the top of the $\cap$ (local max); the green dot at $(1,-2)$ sits at the bottom of the $\cup$ (local min). Notice how the curve is visibly frowning left of $0$ and smiling right of $0$ — that is the sign change of $f^{\prime \prime }$ made visual.

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Example (B) — Cell C4: the $x^4$ trap

Step 1 — critical point. $f^{\prime }(x)=4x^3=0\Rightarrow x=0$. Why? Slope flat only at the origin.

Step 2 — try the second derivative test. $f^{\prime \prime }(x)=12x^2$, so $f^{\prime \prime }(0)=0$. Why? This is the inconclusive case — the test refuses to answer.

Step 3 — fall back to the first-derivative sign test. For $x<0$: $f^{\prime }=4x^3<0$. For $x>0$: $f^{\prime }>0$. Slope goes $-\to +$ ⇒ local minimum. Why? When $f^{\prime \prime }=0$, only neighbour-slopes decide max/min. See First derivative test.

Step 4 — inflection? No. $f^{\prime \prime }(x)=12x^2\ge 0$ everywhere — concave up on both sides. No sign flip ⇒ no inflection, even though $f^{\prime \prime }(0)=0$. Why? $f^{\prime \prime }$ merely touched zero; concavity never changed.

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Example (C) — Cells C3, C6: periodic inflections

Step 1 — concavity tool. $f^{\prime }(x)=\cos x$, $f^{\prime \prime }(x)=-\sin x$. Why? Sign of $-\sin x$ = concavity directly.

Step 2 — read the sign (Cell C6). On $(0,\pi )$: $\sin x>0\Rightarrow f^{\prime \prime }=-\sin x<0\Rightarrow \cap$ (concave down). On $(\pi ,2\pi )$: $\sin x<0\Rightarrow f^{\prime \prime }>0\Rightarrow \cup$ (concave up). Why? We split the domain wherever $\sin x$ changes sign.

Step 3 — inflection (Cell C3). At $x=\pi$: $f^{\prime \prime }=-\sin \pi =0$ and concavity flips $\cap \to \cup$; $f$ continuous ⇒ inflection at $(\pi ,0)$. Why? Sign flip + continuity is the full definition.

Step 4 — the endpoints (edge case under a domain restriction). At $x=0$ and $x=2\pi$ we have $f^{\prime \prime }=-\sin (0)=0$ and $f^{\prime \prime }=-\sin (2\pi )=0$ — both look like candidates. But an inflection needs concavity to change on both sides, and here the domain $[0,2\pi ]$ gives us only one side to look at. At $x=0$ we can only see $x>0$ (concave down); at $x=2\pi$ only $x<2\pi$ (concave up). With no neighbourhood spanning the point, there is no sign change to test, so the endpoints are not inflection points of the restricted function. Why this step? The definition demands concavity switch across the point; a one-sided endpoint cannot switch. (If you extend to all of $\mathbb{R}$, then $x=0$ and $x=2\pi$ do become genuine inflections, because now both sides exist.)

Figure s02 — colour-coded concavity. The red-shaded region on $(0,\pi )$ is where the curve frowns ($f^{\prime \prime }<0$, $\cap$); the green-shaded region on $(\pi ,2\pi )$ is where it smiles ($f^{\prime \prime }>0$, $\cup$). The orange dot marks the single interior inflection at $(\pi ,0)$ where the shading colour flips. Look at the two ends: each touches only one colour, so there is nothing to switch — exactly the endpoint argument above.

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Example (D) — Cell C5: inflection where $f^{\prime \prime }$ does not exist

Step 1 — derivatives. $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3},f^{\prime \prime }(x)=-\frac{2}{9}{x}^{-5/3}=\frac{-2}{9x^{5/3}}.$ Why? Power rule; keep going to $f^{\prime \prime }$ for concavity.

Step 2 — note the gap. At $x=0$, $x^{-5/3}$ blows up: $f^{\prime \prime }(0)$ does not exist (Cell C5). This is still a valid inflection candidate per the definition. Why? Inflection candidates are "$f^{\prime \prime }=0$ or $f^{\prime \prime }$ undefined."

Step 3 — sign check across $0$. For $x>0$: $x^{5/3}>0\Rightarrow f^{\prime \prime }=-\frac{2}{9x^{5/3}}<0\Rightarrow \cap$. For $x<0$: $x^{5/3}<0\Rightarrow f^{\prime \prime }>0\Rightarrow \cup$. Concavity flips $\cup \to \cap$, and $f$ is continuous at $0$ ⇒ inflection at $(0,0)$. Why? Even without $f^{\prime \prime }(0)$, the sign change on either side is what defines an inflection.

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Example (E) — Cell C7: restricted domain / endpoints

Step 1 — interior critical points inside the interval. From (A), $g^{\prime }=0$ at $x=\pm 1$. Only $x=-1$ lies in $[-2,0.5]$ ($x=1$ is outside). Why? The second derivative test classifies interior stationary points only.

Step 2 — classify the interior point. $g^{\prime \prime }(-1)=6(-1)=-6<0\Rightarrow$ local max, value $g(-1)=(-1{)}^3-3(-1)=-1+3=2$. Why? Concave down at a flat point = peak.

Step 3 — check the endpoints (test is silent here). $g(-2)=(-2{)}^3-3(-2)=-8+6=-2,g(0.5)=(0.5{)}^3-3(0.5)=0.125-1.5=-\mathrm{1.375.}$ Why this step? On a closed interval the global min/max may sit at an endpoint where $g^{\prime }\ne 0$ — the second derivative test says nothing about edges, so evaluate directly. See Maxima and minima — optimization.

Step 4 — compare all candidates. Values: $g(-2)=-2$, $g(-1)=2$, $g(0.5)=-1.375$. Global max $=2$ at $x=-1$; global min $=-2$ at the endpoint $x=-2$. Why? The largest/smallest of {critical values, endpoint values} wins.

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Example (F) — Cell C8: real-world optimization word problem

Step 1 — translate words into a function. The base becomes $(12-2x)$ by $(12-2x)$, height $x$: $V(x)=x(12-2x{)}^2,0<x<6.$ Why? Every optimization starts by writing the quantity to maximize as one function of one variable, with a physical domain. See Curve sketching.

Step 2 — expand and differentiate. $V(x)=x(144-48x+4x^2)=144x-48x^2+4x^3,$ $V^{\prime }(x)=144-96x+12x^2=12(x^2-8x+12)=12(x-2)(x-6).$ Why? Set $V^{\prime }=0$ to find candidate cut sizes.

Step 3 — critical points. $V^{\prime }=0\Rightarrow x=2$ or $x=6$. Only $x=2$ is inside $(0,6)$. Why? $x=6$ collapses the base to zero area — not a real box.

Step 4 — classify with the second derivative test (Cell C8). $V^{\prime \prime }(x)=-96+24x,V^{\prime \prime }(2)=-96+48=-48<0\Rightarrow \cap \Rightarrow \text{maximum}.$ Why? Concave down at the flat point = top of the volume hill — this proves it's a max, not a min.

Step 5 — the answer with units. $V(2)=2(12-4{)}^2=2\cdot 64=128{\text{ cm}}^3.$ Why? Plug the winning $x$ back in; volume carries ${\text{cm}}^3$.

Figure s03 — the volume hill. The blue curve is $V(x)$ over the physical range $0<x<6$. The red dot marks the single peak at $x=2$, $V=128{\text{cm}}^3$; the curve is concave down ($\cap$) there, which is why the second derivative test returns "maximum". Cutting less or more than $2$ cm slides you down either side of the hill.

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Example (G) — Cell C9: exam twist with an unknown parameter

Step 1 — the tool. $f^{\prime }(x)=3x^2+2ax+1$, $f^{\prime \prime }(x)=6x+2a$. Why? Inflection is about $f^{\prime \prime }$, so build it.

Step 2 — force the candidate condition at $x=1$. $f^{\prime \prime }(1)=6(1)+2a=0\Rightarrow 6+2a=0\Rightarrow a=-3.$ Why? An inflection requires $f^{\prime \prime }(c)=0$ (or undefined); here we solve for the $a$ that makes it so.

Step 3 — confirm the sign actually flips (Cell C9, the trap-guard). With $a=-3$: $f^{\prime \prime }(x)=6x-6=6(x-1)$. For $x<1$: $f^{\prime \prime }<0$ ($\cap$). For $x>1$: $f^{\prime \prime }>0$ ($\cup$). It flips ⇒ genuine inflection at $x=1$. Why? Same lesson as (B): $f^{\prime \prime }=0$ alone is not enough. Since $f^{\prime \prime }$ is a straight line crossing zero, the sign genuinely changes — the twist is safely resolved.

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Coverage check

Recall Did we hit every matrix cell?

C1 min → (A). C2 max → (A). C3 real inflection → (A),(C). C4 fake ($x^4$) → (B). C5 $f^{\prime \prime }$ undefined → (D). C6 periodic → (C). C7 endpoints → (E). C8 word problem → (F). C9 parameter twist → (G). ✓ All nine.

Connections

• Second derivative test — concavity, inflection points
• First derivative test
• Critical points
• Taylor series
• Curve sketching
• Maxima and minima — optimization
• Concavity and convex functions