Worked examples — Second derivative test — concavity, inflection points

4.1.29 · D3 · Maths › Calculus I — Limits & Derivatives › Second derivative test — concavity, inflection points

Yeh page ek drill hall hai. Parent note ne tools banaye; yahaan hum har tarah ke questions unpe fire karte hain. Kuch bhi solve karne se pehle, hum ek matrix banate hain — is topic ke har possible case class ki — phir har worked example ko exactly usi cell se tag kiya gaya hai jise woh cover karta hai, taaki end tak koi bhi scenario aisa na rahe jo tumne na dekha ho.

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Scenario matrix

Is topic ka har problem inhi cells mein se kisi ek mein aata hai. Baad ke worked examples cell ke naam se label hain taaki tum dekh sako ki coverage complete hai.

Cell	Case class	Tricky kyun hai	Example
C1	$f^{\prime \prime }(c)>0$ critical point par	clean minimum	(A)
C2	$f^{\prime \prime }(c)<0$ critical point par	clean maximum	(A)
C3	$f^{\prime \prime }(c)=0$, sign flip hoti hai	genuine inflection	(A), (C)
C4	$f^{\prime \prime }(c)=0$, sign flip nahi hoti	trap: inflection nahi, test fail	(B)
C5	$c$ par $f^{\prime \prime }$ exist nahi karta	corner/cusp wala inflection $f^{\prime \prime }$ mein	(D)
C6	Multiple/periodic inflections	trig, infinitely many switches	(C)
C7	Endpoint / restricted domain	2nd-deriv test edges par kuch nahi kehta	(E)
C8	Real-world optimization word problem	words → $f$ translate karo, phir classify karo	(F)
C9	Exam twist: unknown parameter	ek constant choose karo taaki ek point inflection ho	(G)

Hum saat examples mein saari nau cells cover karenge.

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Example (A) — Cells C1, C2, C3: ek cubic par poora sweep

Step 1 — road flat kahan hoti hai yeh dhundho. $f^{\prime }(x)=3x^2-3=0\Rightarrow x^2=1\Rightarrow x=\pm 1.$ Yeh step kyun? Critical points ($f^{\prime }=0$) hi woh jagahein hain jahaan local max ya min chhup sakta hai — wahan tangent horizontal hoti hai.

Step 2 — concavity tool lao. $f^{\prime \prime }(x)=6x.$ Yeh step kyun? Ek flat point par, $f^{\prime \prime }$ ka sign cup vs cap decide karta hai, isliye min vs max — neighbours test karne ki zaroorat nahi.

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

Step 4 — inflection candidate (Cell C3). $f^{\prime \prime }(x)=6x=0\Rightarrow x=0.$ Sign check: $x<0$ ke liye, $f^{\prime \prime }<0$ ($\cap$); $x>0$ ke liye, $f^{\prime \prime }>0$ ($\cup$). Yeh flip hoti hai, toh $(0,f(0))=(0,0)$ ek genuine inflection hai. Yeh step kyun? $f^{\prime \prime }=0$ sirf ek candidate hai; concavity ka actually switch karna zaroori hai.

Figure s01 — poori cubic ek nazar mein. Neeche ka plot teen saari features ek curve par dikhata hai: $(0,0)$ par orange dot woh inflection hai jahaan road frown ($\cap$) se smile ($\cup$) mein switch hoti hai; $(-1,2)$ par red dot $\cap$ ke top par hai (local max); $(1,-2)$ par green dot $\cup$ ke bottom par hai (local min). Notice karo ki curve $0$ ke left mein visibly frown kar rahi hai aur $0$ ke right mein smile — yahi $f^{\prime \prime }$ ka sign change visually hai.

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Example (B) — Cell C4: $x^4$ ka trap

Step 1 — critical point. $f^{\prime }(x)=4x^3=0\Rightarrow x=0$. Kyun? Slope sirf origin par flat hai.

Step 2 — second derivative test try karo. $f^{\prime \prime }(x)=12x^2$, toh $f^{\prime \prime }(0)=0$. Kyun? Yeh inconclusive case hai — test answer dene se mana karta hai.

Step 3 — first-derivative sign test par fall back karo. $x<0$ ke liye: $f^{\prime }=4x^3<0$. $x>0$ ke liye: $f^{\prime }>0$. Slope $-\to +$ jaati hai ⇒ local minimum. Kyun? Jab $f^{\prime \prime }=0$ ho, toh sirf neighbour-slopes max/min decide karte hain. Dekho First derivative test.

Step 4 — inflection? Nahi. $f^{\prime \prime }(x)=12x^2\ge 0$ har jagah — dono taraf concave up. Sign flip nahi ⇒ inflection nahi, chahe $f^{\prime \prime }(0)=0$ ho. Kyun? $f^{\prime \prime }$ ne sirf zero touch kiya; concavity kabhi badi nahi.

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

Step 1 — concavity tool. $f^{\prime }(x)=\cos x$, $f^{\prime \prime }(x)=-\sin x$. Kyun? $-\sin x$ ka sign directly concavity deta hai.

Step 2 — sign padho (Cell C6). $(0,\pi )$ par: $\sin x>0\Rightarrow f^{\prime \prime }=-\sin x<0\Rightarrow \cap$ (concave down). $(\pi ,2\pi )$ par: $\sin x<0\Rightarrow f^{\prime \prime }>0\Rightarrow \cup$ (concave up). Kyun? Hum domain ko wahan split karte hain jahaan $\sin x$ sign change karta hai.

Step 3 — inflection (Cell C3). $x=\pi$ par: $f^{\prime \prime }=-\sin \pi =0$ aur concavity $\cap \to \cup$ flip hoti hai; $f$ continuous hai ⇒ inflection at $(\pi ,0)$. Kyun? Sign flip + continuity — yahi poori definition hai.

Step 4 — endpoints (domain restriction ka edge case). $x=0$ aur $x=2\pi$ par hame $f^{\prime \prime }=-\sin (0)=0$ aur $f^{\prime \prime }=-\sin (2\pi )=0$ milta hai — dono candidates lagte hain. Lekin ek inflection ke liye concavity ko dono taraf change karna chahiye, aur yahaan domain $[0,2\pi ]$ hamein sirf ek side deta hai. $x=0$ par hum sirf $x>0$ dekh sakte hain (concave down); $x=2\pi$ par sirf $x<2\pi$ (concave up). Jab koi neighbourhood point ko span nahi karti, toh koi sign change test karne ke liye nahi hai, isliye endpoints restricted function ke inflection points nahi hain. Yeh step kyun? Definition maangti hai ki concavity across the point switch kare; ek one-sided endpoint switch nahi kar sakta. (Agar $\mathbb{R}$ tak extend karo, toh $x=0$ aur $x=2\pi$ genuine inflections ban jaate hain, kyunki ab dono sides exist karti hain.)

Figure s02 — colour-coded concavity. $(0,\pi )$ par red-shaded region woh jagah hai jahaan curve frown karta hai ($f^{\prime \prime }<0$, $\cap$); $(\pi ,2\pi )$ par green-shaded region woh jagah hai jahaan smile karta hai ($f^{\prime \prime }>0$, $\cup$). Orange dot $(\pi ,0)$ par single interior inflection mark karta hai jahaan shading color flip hoti hai. Do ends dekho: har ek sirf ek colour touch karta hai, toh switch karne ke liye kuch nahi — yahi upar wala endpoint argument hai.

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Example (D) — Cell C5: inflection jahaan $f^{\prime \prime }$ exist nahi karta

Step 1 — derivatives. $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3},f^{\prime \prime }(x)=-\frac{2}{9}{x}^{-5/3}=\frac{-2}{9x^{5/3}}.$ Kyun? Power rule; concavity ke liye $f^{\prime \prime }$ tak jaate raho.

Step 2 — gap note karo. $x=0$ par, $x^{-5/3}$ blow up karta hai: $f^{\prime \prime }(0)$ exist nahi karta (Cell C5). Phir bhi yeh definition ke hisaab se valid inflection candidate hai. Kyun? Inflection candidates hain "$f^{\prime \prime }=0$ ya $f^{\prime \prime }$ undefined."

Step 3 — $0$ ke aas-paas sign check karo. $x>0$ ke liye: $x^{5/3}>0\Rightarrow f^{\prime \prime }=-\frac{2}{9x^{5/3}}<0\Rightarrow \cap$. $x<0$ ke liye: $x^{5/3}<0\Rightarrow f^{\prime \prime }>0\Rightarrow \cup$. Concavity $\cup \to \cap$ flip hoti hai, aur $f$ $0$ par continuous hai ⇒ inflection at $(0,0)$. Kyun? $f^{\prime \prime }(0)$ ke bina bhi, dono taraf sign change hi inflection define karta hai.

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

Step 1 — interval ke andar interior critical points. (A) se, $g^{\prime }=0$ at $x=\pm 1$. Sirf $x=-1$, $[-2,0.5]$ mein hai ($x=1$ bahar hai). Kyun? Second derivative test sirf interior stationary points classify karta hai.

Step 2 — interior point classify karo. $g^{\prime \prime }(-1)=6(-1)=-6<0\Rightarrow$ local max, value $g(-1)=(-1{)}^3-3(-1)=-1+3=2$. Kyun? Flat point par concave down = peak.

Step 3 — endpoints check karo (test yahaan chup hai). $g(-2)=(-2{)}^3-3(-2)=-8+6=-2,g(0.5)=(0.5{)}^3-3(0.5)=0.125-1.5=-\mathrm{1.375.}$ Yeh step kyun? Closed interval par global min/max ek endpoint par baith sakta hai jahaan $g^{\prime }\ne 0$ — second derivative test edges ke baare mein kuch nahi kehta, toh directly evaluate karo. Dekho Maxima and minima — optimization.

Step 4 — saare candidates compare karo. Values: $g(-2)=-2$, $g(-1)=2$, $g(0.5)=-1.375$. Global max $=2$ at $x=-1$; global min $=-2$ at endpoint $x=-2$. Kyun? {critical values, endpoint values} mein se sabse bada/chota jeetta hai.

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

Step 1 — words ko function mein translate karo. Base $(12-2x)$ by $(12-2x)$ ban jaata hai, height $x$: $V(x)=x(12-2x{)}^2,0<x<6.$ Kyun? Har optimization ek quantity ko maximize karne ke liye ek variable ka ek function likhne se shuru hota hai, physical domain ke saath. Dekho Curve sketching.

Step 2 — expand karo aur differentiate karo. $V(x)=x(144-48x+4x^2)=144x-48x^2+4x^3,$ $V^{\prime }(x)=144-96x+12x^2=12(x^2-8x+12)=12(x-2)(x-6).$ Kyun? $V^{\prime }=0$ set karo candidate cut sizes dhundne ke liye.

Step 3 — critical points. $V^{\prime }=0\Rightarrow x=2$ ya $x=6$. Sirf $x=2$, $(0,6)$ ke andar hai. Kyun? $x=6$ base area zero kar deta hai — real box nahi banta.

Step 4 — second derivative test se classify karo (Cell C8). $V^{\prime \prime }(x)=-96+24x,V^{\prime \prime }(2)=-96+48=-48<0\Rightarrow \cap \Rightarrow \text{maximum}.$ Kyun? Flat point par concave down = volume hill ka top — yeh prove karta hai ki yeh max hai, min nahi.

Step 5 — units ke saath answer. $V(2)=2(12-4{)}^2=2\cdot 64=128{\text{ cm}}^3.$ Kyun? Winning $x$ wapas plug karo; volume ${\text{cm}}^3$ mein hota hai.

Figure s03 — volume hill. Blue curve physical range $0<x<6$ par $V(x)$ hai. Red dot $x=2$, $V=128{\text{cm}}^3$ par single peak mark karta hai; curve wahan concave down ($\cap$) hai, yahi wajah hai ki second derivative test "maximum" return karta hai. $2$ cm se kam ya zyada kaatna tumhe hill ke dono taraf neeche le jaata hai.

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

Step 1 — tool banao. $f^{\prime }(x)=3x^2+2ax+1$, $f^{\prime \prime }(x)=6x+2a$. Kyun? Inflection $f^{\prime \prime }$ ke baare mein hai, toh pehle use banao.

Step 2 — $x=1$ par candidate condition force karo. $f^{\prime \prime }(1)=6(1)+2a=0\Rightarrow 6+2a=0\Rightarrow a=-3.$ Kyun? Inflection ke liye $f^{\prime \prime }(c)=0$ (ya undefined) chahiye; yahaan hum woh $a$ solve karte hain jo yeh possible banata hai.

Step 3 — confirm karo ki sign actually flip hoti hai (Cell C9, trap-guard). $a=-3$ ke saath: $f^{\prime \prime }(x)=6x-6=6(x-1)$. $x<1$ ke liye: $f^{\prime \prime }<0$ ($\cap$). $x>1$ ke liye: $f^{\prime \prime }>0$ ($\cup$). Flip hoti hai ⇒ $x=1$ par genuine inflection. Kyun? (B) wali same lesson: $f^{\prime \prime }=0$ akela kaafi nahi hai. Kyunki $f^{\prime \prime }$ ek straight line hai jo zero cross karta hai, sign genuinely change hoti hai — twist safely resolve ho gayi.

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Coverage check

Recall Kya humne har matrix cell hit kiya?

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). ✓ Saari nau.

Connections

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