Second derivative test — concavity, inflection points: Visual walkthrough

4.1.29 · D2 · Maths › Calculus I — Limits & Derivatives › Second derivative test — concavity, inflection points

Koi symbol aane se pehle: ek function $f$ ek rule hai jo ek number $x$ leta hai aur height $f (x)$ deta hai. Har $x$ ke liye point $(x, f (x))$ plot karo aur tumhe ek curve milega — ek number line ke upar bani hui ek road. Wahi road hum dekhenge.

Step 1 — Road ka slope: $f^{'}$ kya measure karta hai

KYA. Curve par ek point pick karo aur ek seedha ruler wahan rakh do jo curve ko bas graze kare — touch kare lekin cross na kare. Woh ruler tangent line hai. Uski steepness — yaani right ki taraf ek step lene par kitni height milti hai — woh slope hai. Is slope ko hum $f^{'} (x)$ kehte hain (padho " $f$ -prime of $x$ "). Chhota sa prime mark $^{'}$ sirf notation hai "slope function of $f$ " ke liye.

YEH TOOL KYUN. Hum bending ki baat karna chahte hain, lekin bending ek change in direction hai, aur direction exactly wahi hai jo slope capture karta hai. Isliye honest starting point slope hi hai. Abhi "second" ke baare mein kuch nahi — pehle hum first mein fluent hona chahte hain.

PICTURE. Ek curve par teen tangent rulers.

Figure — Second derivative test — concavity, inflection points

Step 2 — Cup walk karte waqt slope ko change hote dekho

KYA. Ek bowl-shaped ( $\cup$ ) curve lo aur left-to-right chalte hue har jagah tangent slope padho. Left par ruler steeply down point karta hai (bahut negative slope). Bottom ke paas flat ho jaata hai (zero). Right par up tilt ho jaata hai (positive). Toh slope-number $- 3 \to - 1 \to 0 \to + 1 \to + 3$ chalta hai: yeh increasing hai.

KYUN. Yeh key observable hai. "Concave up" koi nayi mysterious cheez nahi hai — yeh literally slope ka right move karne par bada hona hai. Hum ek shape-word ("cup") ko ek number-behaviour ("slope increasing") mein badal rahe hain.

PICTURE. Wahi bowl jisme paanch jagah tangent rulers banaye hain; slope values labelled hain.

Step 3 — "Slope increasing" measure karo — derivative dobara apply karo

KYA. Hamare paas ek number hai jo $x$ ke chalte waqt change hota hai: slope $f^{'} (x)$ . "Kya yeh number badh raha hai?" poochne ke liye hum exactly wahi karte hain jo Step 1 mein kiya tha — iska slope lete hain. $f^{'}$ ka slope $f^{''} (x)$ likha jaata hai (do primes = "slope of the slope").

YEH TOOL HI KYUN, KUCH AUR KYUN NAHI. Hamare paas pehle se ek perfect device hai "kya koi quantity badh rahi hai?" ke liye: derivative, kyunki $g^{'} > 0$ ka matlab hai " $g$ , $x$ ke badhne par badhta hai." Hume kuch naya invent nahi karna — bas $f^{'}$ ko usi machine mein daalna hai. Yahan derivative choose karna sabse natural choice hai kyunki kisi bhi function ka increase/decrease wahi derivative report karta hai.

PICTURE. Do stacked plots — upar $f^{'} (x)$ hai (hamare cup ke liye ek rising line); neeche uska apna slope $f^{''} (x)$ hai, zero ke upar baitha hua.

Step 4 — Cap ( $\cap$ ): har sign flip ho jaata hai

KYA. Ab ek dome-shaped ( $\cap$ ) curve par chalo. Tangents left par steeply up se, peak par flat hote hue, right par steeply down tak jaate hain: slope $+ 3 \to + 1 \to 0 \to - 1 \to - 3$ chalta hai — decreasing. Slope decreasing matlab uska slope negative hai: $f^{''} < 0$ .

YEH ALAG KYUN DIKHATE HAIN. Contract yeh hai ki har case cover kiya jaaye. Cup aur cap do concavities hain; hume sign genuinely reverse hote dekhna hai, assume nahi karna.

PICTURE. Dome jisme paanch tangent rulers hain, slopes labelled decreasing hain; ek small inset $f^{'}$ ko ek falling line ke roop mein dikhata hai.

Step 5 — Jahan cup, cap banta hai: inflection point

KYA. Ek single curve ko $\cup$ region se $\cap$ region mein slide karo. Kahin na kahin bending reverse hogi. Us switch par slope $f^{'}$ badhna band karke ghata shuru ho jaata hai — toh $f^{''}$ $+$ se $0$ hote hue $-$ mein jaata hai. Woh crossing point $(c, f (c))$ inflection point hai.

$f^{''} = 0$ sirf ek clue kyun hai. $f^{''}$ ke $+$ se $-$ jaane ke liye agar woh continuous hai toh use zero cross karna hi hoga. Lekin zero touch karna usse cross karna nahi hai — ek value $0$ ko touch karke usi sign mein wapas bounce kar sakti hai. Isliye definition mein genuinely sign change demand ki jaati hai, sirf $f^{''} (c) = 0$ nahi.

PICTURE. Ek S-shaped curve; left half $\cup$ shade ki hui, right $\cap$ , join inflection mark hai, neeche $f^{''}$ sign strip $+ 0 -$ hai.

Step 6 — Degenerate trap: $x^{4}$ zero ko touch karta hai lekin kabhi palat nahi

KYA. Lo $f (x) = x^{4}$ . Iska slope-of-slope $f^{''} (x) = 12 x^{2}$ hai. $x = 0$ par yeh $0$ ke barabar hai — "clue" fire karta hai. Lekin $12 x^{2} \geq 0$ har $x$ ke liye hai (square kabhi negative nahi hota). Toh $f^{''}$ $0$ tak girta hai aur sign change kiye bina wapas utha jaata hai: dono taraf concave up. Koi inflection nahi, $f^{''} (0) = 0$ hone ke baawajood.

YEH KYUN MAAYने RAKHTA HAI. Yeh woh case hai jise naive rule " $f^{''} = 0 \Rightarrow$ inflection" galat padhti hai. $12 x^{2}$ ko axis kiss karte dekh ke non-negative rehna hi poora lesson hai.

PICTURE. Upar: flat-bottomed $x^{4}$ bowl. Neeche: $f^{''} = 12 x^{2}$ jo $0$ par axis touch karta hai lekin uske upar rehta hai.

Step 7 — Doosra degenerate case: inflection jahan $f^{''}$ exist hi nahi karta

KYA. Definition ne ek doosra door allow kiya tha: ek inflection jahan $f^{''} (c)$ exist karna hi band kar de. Yeh raha honest example — cube root $f (x) = x^{1/3}$ . Do baar differentiate karo: $f^{'} (x) = \frac{1}{3} x^{- 2/3}, f^{''} (x) = constant - \frac{2}{9} = 1/ x^{5/3} x^{- 5/3} .$ $x = 0$ par denominator $x^{5/3}$ zero hai, isliye $f^{''} (0)$ undefined hai — division by zero, curve wahan vertical hai. Lekin har side ka sign dekho: $x < 0$ ke liye, $x^{5/3} < 0$ toh $f^{''} = - \frac{2}{9} \cdot (neg) > 0$ (concave up, $\cup$ ); $x > 0$ ke liye, $x^{5/3} > 0$ toh $f^{''} < 0$ (concave down, $\cap$ ).

YEH KYUN MAAYने RAKHTA HAI. Concavity genuinely $x = 0$ ke across $\cup \to \cap$ flip karti hai, aur $f$ wahan continuous hai, isliye $(0, 0)$ sach mein ek inflection point hai — chahe $f^{''}$ kabhi zero nahi hua aur switch par defined bhi nahi tha. Yeh definition ka doosra door band karta hai jo Step 6 ne khola tha.

PICTURE. Upar: vertical-tangent cube-root curve, pehle up-bending phir down-bending. Neeche: $f^{''}$ left se $+ \infty$ aur right se $- \infty$ tak jaata hai — ek genuine sign change through a break, zero se nahi.

Step 8 — Payoff: concavity ek flat point classify kyun karti hai (Taylor)

KYA. Ek critical point $c$ par tangent flat hai: $f^{'} (c) = 0$ (yahi Critical points ka ghar hai). $c$ ke paas, Taylor's expansion height difference ko exactly bending term aur leftover ke roop mein likhti hai: $f (x) - f (c) = constant \frac{1}{2} concavity sign f^{''} (c) \geq 0 always (x - c)^{2} + remainder R (x), \frac{R ( x )}{( x - c ) ^{2}} x \to c 0.$

YEH TOOL KYUN. Hum nearby $x$ ke liye height $f (x)$ ko flat-point height $f (c)$ se compare karna chahte hain. Taylor's formula woh microscope hi hai jo us difference ko $f^{''}$ ke terms mein likhta hai. Linear term is liye mara kyunki $f^{'} (c) = 0$ , aur square term charge mein aa gaya.

Sign padho. Chunki $(x - c)^{2} \geq 0$ hamesha hai aur $R$ kaafi close to $c$ par negligible hai (upar wala limit), difference $f (x) - f (c)$ ka sign $f^{''} (c)$ ka sign hai:

$f^{''} (c) > 0 \Rightarrow f (x) > f (c)$ nearby ⇒ valley mein baithe ho ⇒ local minimum.
$f^{''} (c) < 0 \Rightarrow f (x) < f (c)$ nearby ⇒ peak par baithe ho ⇒ local maximum.
$f^{''} (c) = 0 \Rightarrow$ square term vanish ⇒ $R$ ab decide karega ⇒ inconclusive, First derivative test par jaao.

PICTURE. Ek cup jisme base par flat tangent hai (min) aur ek cap jisme crest par flat tangent hai (max); parabola $\frac{1}{2} f^{''} (c) (x - c)^{2}$ dono ke upar drawn hai.

Ek-picture summary

Sab kuch compress kiya: upar curve, beech mein slope $f^{'}$ , neeche concavity $f^{''}$ — teen plots ek $x$ -axis share karte hain taaki tum poori kahaani vertically padh sako.

Recall Feynman retelling — poora walk simple words mein

Ek pahadi road banao. Slope batata hai ki tum upar ja rahe ho ya neeche (woh hai $f^{'}$ ). Ab height mat dekho — chalte waqt slope khud dekho. Agar slope hamesha barhta rehta hai (steep-down se, flat se hote hue, steep-up tak), toh road smile ki tarah curve karti hai, ek cup: woh hai $f^{''} > 0$ . Agar slope hamesha ghatta rehta hai, toh road frown karti hai, ek cap: $f^{''} < 0$ . "Kya slope badh raha hai?" measure karna wahi kaam hai jo slope measure karna hai, isliye hum slope ka slope le lete hain — woh hain do chhote primes. Jahan smile frown mein badal jaati hai, slope badhna band karke ghatta shuru ho jaata hai, toh $f^{''}$ sign flip karta hai — woh hai inflection point — ya toh zero se hoke (jaisa $x^{3}$ ) ya ek break se hoke jahan $f^{''}$ blow up karta hai (jaisa cube root). Yeh inflection nahi hai agar road sirf flatness kiss kare aur same curve rakhhe (jaisa $x^{4}$ ). Aakhir mein, agar tum ek flat spot par ruko: cup-flat ek valley ka bottom hai (min), cap-flat ek bump ka top hai (max). Yahi poora second-derivative test hai, dekha hua, yaad nahi kiya hua.

Recall Quick self-check (koshish ke baad kholo)

Q1. $f^{''} > 0$ geometrically kya matlab hai? → Concave up ( $\cup$ ); slope $f^{'}$ increasing hai. Q2. Concavity ke liye derivative do baar lena sahi tool kyun hai? → Concavity poochhti hai "kya slope badh raha hai?", aur derivative exactly "kya badh raha hai?" ka detector hai — $f^{'}$ par apply kiya gaya. Q3. Kya $f^{''} (c) = 0$ inflection guarantee karne ke liye kaafi hai? → Nahi; $f^{''}$ ka sign actually change hona chahiye. $x^{4}$ mein $f^{''} = 12 x^{2} \geq 0$ hai, koi flip nahi. Q4. Kya inflection wahan bhi ho sakta hai jahan $f^{''}$ exist na kare? → Haan; $x^{1/3}$ at $0$ $\cup \to \cap$ flip karta hai jab $f^{''} (0)$ undefined hai. Q5. $f^{'} (c) = 0$ aur $f^{''} (c) < 0$ — classify karo. → Local maximum (cap-flat = peak).

Visual walkthrough — Second derivative test — concavity, inflection points