Second derivative test — concavity, inflection points: Question bank

4.1.29 · D5 · Maths › Calculus I — Limits & Derivatives › Second derivative test — concavity, inflection points

Intuition Is bank ko kaise use karein

Neeche har ek item ek conceptual trap hai: ek aisi jagah jahan " $f^{''} > 0$ ", " $f^{''} = 0$ ", "inflection", ya "concave" jaisi words tumhe ek galat shortcut ki taraf kheenchti hain. Har line ko reveal karne se pehle zor se bol ke dekho — sirf haan/nahi nahi, balki kyun bolo. Yeh parent test ko uske apne edge cases ke against stress-test kar raha hai.

Un vocabulary terms ki yaad dilata hoon jinka tum sahara loge:

$f^{'}$ = slope (upar/neeche).
$f^{''}$ = slope ka slope → concavity ( $+$ matlab cup $\cup$ , $-$ matlab cap $\cap$ ).
Critical point: woh jagah jahan $f^{'} = 0$ ya $f^{'}$ undefined ho (flat tangent, corner, ya cusp) — dekho Critical points. First-derivative test dono tarah ke points classify karta hai.
Inflection point: ek continuous jagah jahan concavity ka sign palat jaata hai.

Definition Strict vs. non-strict concavity (ek wording caution)

"Concave up" words ke peeche do precise notions chhupe hain, aur neeche ke traps usi gap ko exploit karte hain:

Strictly concave up kisi interval par matlab hai ki $f^{''} > 0$ poore interval mein (slope strictly increasing). Yahi sense is bank mein har jagah use hota hai jab tak koi item aur na kahe.
Non-strictly (weakly) concave up matlab hai $f^{''} \geq 0$ , yani $f^{''}$ isolated points par zero ko chhoo sakta hai lekin kabhi negative nahi jaata.

Yeh distinction kyun matter karta hai: $f (x) = x^{4}$ ke liye hum kehte hain "yeh $0$ ke dono taraf concave up rehta hai", jo sirf non-strict $f^{''} \geq 0$ sense mein sach hai — single point $x = 0$ par $f^{''} = 0$ hai, isliye wahan yeh strictly concave up nahi hai. Kyunki $f^{''}$ kabhi negative nahi hota, concavity kabhi flip nahi hoti, isliye abhi bhi no inflection hai. Dono readings ko alag rakho aur $x^{4}$ wala trap khud hi gayab ho jaata hai.

Neeche ke teen blueprint figures sirf decoration nahi hain — har ek figure ek specific misconception settle karta hai jise reveals baar baar test karte hain. Matching group tackle karne se pehle har figure ke neeche ka caption note zaroor padho.

Figure 1 — $f^{''}$ ka sign hi concavity hai. Cyan curve ( $f^{''} > 0$ ) paani rok ke rakhta hai jaise ek cup; amber curve ( $f^{''} < 0$ ) paani baha deta hai jaise ek cap. "True or false" sign items ka poora content yahi hai: jab bhi tum " $f^{''} > 0$ " padho, cyan cup ki picture banao, aur "up is a cuP → min" sirf ek slogan nahi rehta.

Figure — Second derivative test — concavity, inflection points

Figure 2 — zero cross karna vs. zero ko chhona. Cyan line $f^{''} = 6 x$ (from $x^{3}$ ) axis ko cross karti hai, isliye concavity genuinely flip hoti hai → inflection. Amber parabola $f^{''} = 12 x^{2}$ (from $x^{4}$ ) sirf chhoo ke waapas non-negative ho jaati hai → koi flip nahi, koi inflection nahi. Bank mein har " $f^{''} = 0$ therefore inflection" trap is ek picture par jeeta hai.

Figure 3 — flat point par, concavity akele decide karti hai. Dono curves ka horizontal tangent hai (dashed white). Cyan cup us flat point ko bottom par rakhta hai (minimum); amber cap use top par rakhta hai (maximum). Isliye second derivative test ko pehle $f^{'} (c) = 0$ chahiye — "Spot the error" ke saare sign confusions isi wajah se hain kyunki log bhool jaate hain ki flat point kis cup mein baitha hai.

True or false — justify

Har ek ek claim hai. Reveal mein True/False aur woh reasoning diya gaya hai jo use settle karta hai.

Agar

f^{''} (c) = 0

toh

c

ek inflection point hai.

False.

f^{''}

c

par sign change karna chahiye;

f (x) = x^{4}

f^{''} (0) = 0

hai lekin

f^{''} = 12 x^{2} \geq 0

dono taraf hai, isliye yeh concave up rehta hai — koi flip nahi, koi inflection nahi.

Agar

f

kisi interval par concave up hai, toh

f

wahan increasing hai.

False. Concave up ka matlab hai slope

f^{'}

increasing hai,

f

khud nahi.

f (x) = x^{2}

(- \infty, 0)

concave up hai lekin decreasing hai.

Test se mila local minimum

f^{''} (c)

strictly positive hona chahiye.

False. Minimum ka

f^{''} (c) = 0

bhi ho sakta hai; e.g.

x^{4}

0

ek genuine minimum hai jahan

f^{''} (0) = 0

aur test sirf inconclusive hai, galat nahi.

Agar

f^{''} > 0

har jagah ho AUR kisi

c

par

f^{'} (c) = 0

ho, toh woh critical point ek minimum hai, kabhi maximum nahi.

True. Local extremum hone ke liye pehle ek critical point chahiye; ek milne par,

f^{''} > 0

globally concave up (

\cup

) dikhata hai, isliye flat point cup ka bottom hai — ek minimum. Local maximum ke liye kahin concave-down behaviour chahiye hoga, jo

f^{''} > 0

forbid karta hai.

Inflection point hamesha kisi critical point par hona chahiye.

False. Inflection

f^{''}

ke flip hone ke baare mein hai, critical

f^{'} = 0

(ya undefined) ke baare mein.

sin x

x = π

par inflect karta hai, jahan

f^{'} = cos π = - 1 \neq = 0

— flat slope zaroori nahi.

Agar

f^{''} (c)

exist nahi karta, toh

c

inflection point nahi ho sakta.

False. Concavity tab bhi switch ho sakti hai jahan

f^{''}

undefined ho; e.g.

f (x) = x^{1/3}

0

par inflect karta hai jahan

f^{''}

blow up karta hai, aur graph

\cup \to \cap

jaata hai.

Ek function ek hi point par concave up aur concave down dono ho sakta hai.

False. Ek single point par concavity ek sign ki hoti hai (ya undefined); "concavity change hoti hai" ka matlab hai point ke dono sides ke baare mein, point khud ke baare mein nahi.

Agar

f^{''} (c) > 0

toh

f

c

par minimum hai.

Jaisa likha hai, False — tumhe pehle jaanna chahiye ki

f^{'} (c) = 0

f^{''} > 0

non-critical point par sirf dikhata hai ki graph wahan cup ki tarah curve karta hai, kisi bhi slope ke saath.

Do consecutive inflection points ke beech, concavity constant rehti hai.

True. Inflections exactly

f^{''}

ke sign changes hain; dono ke beech

f^{''}

ek hi sign rakhta hai, isliye concavity switch nahi hoti.

Ek straight line

f (x) = m x + b

har jagah inflection point hai.

False.

f^{''} \equiv 0

hai lekin yeh kabhi sign nahi badalta (hamesha zero rehta hai), isliye concavity kabhi flip nahi hoti — koi inflection points nahi hain.

Spot the error

Har line mein ek flawed argument hai. Reveal exactly woh broken step batata hai.

x^{4}

ke liye

f^{''} (0) = 0

hai, isliye second derivative test se

x = 0

na max hai na min."

Error yeh hai:

f^{''} = 0

test ko inconclusive banata hai, yeh "neither" declare nahi karta. First-derivative test dikhata hai ki

f^{'}

- \to +

jaata hai, isliye

0

ek genuine minimum hai. (Yeh bhi note karo: "saddle" higher dimensions ke liye reserved hai — 1D mein "neither max nor min" bolo.)

f^{'} (c) = 0

aur

f^{''} (c) < 0

, isliye

c

par graph sabse neeche hai."

Sign confusion:

f^{''} (c) < 0

matlab concave down (

\cap

), jo ek maximum deta hai (nearby sabse upar), na ki sabse neeche. Yaad rakho "up is a cuP → min".

"Concavity

x = c

par change hoti hai, aur

f

wahan jump karta hai, isliye yeh inflection point hai."

Inflection ke liye

f

c

par continuous hona chahiye. Jump isko disqualify karta hai, bhaley hi concavity technically dono sides par alag ho.

f^{''}

c

par positive hai, isliye tangent line

c

ke paas curve ke upar hai."

Ulta hai: concave up (

\cup

) matlab curve tangent se upar ki taraf bend karta hai, isliye tangent curve ke neeche hoti hai paas mein.

"Humne

f^{''} (c) = 0

paaya, isliye sign changes check karna skip karte hain aur conclude karte hain ki koi inflection nahi."

Sign check skip karna hi poora trap hai.

f^{''} (c) = 0

sirf ek candidate hai; kuch candidates (jaise

x^{3}

0

) sign flip karte hain aur hain inflections.

"Second derivative test

c

par fail hua, isliye

c

ek extremum nahi hai."

"Test fail" ka matlab sirf yeh hai ki usne koi verdict nahi diya. Tumhe First derivative test par fall back karna hoga; point abhi bhi max, min, ya neither ho sakta hai.

Why questions

Sirf fact nahi, reasoning explain karo.

Taylor argument mein

(x - c)^{2}

ka sign conclusion ko kabhi affect kyun nahi karta?

Kyunki

(x - c)^{2} \geq 0

hamesha hota hai, isliye

f (x) - f (c) \approx \frac{1}{2} f^{''} (c) (x - c)^{2}

ka sign sirf

f^{''} (c)

se fix hota hai — yahi poori wajah hai ki

f^{''}

min vs max decide karta hai (dekho Taylor series).

Second derivative test

x^{4}

ke critical point ko kabhi classify kyun nahi kar sakta?

Saare lower Taylor terms vanish ho jaate hain:

f^{'} (0) = f^{''} (0) = f^{'''} (0) = 0

aur pehla nonzero term

x^{4}

hai. Test sirf

f^{''}

(quadratic) term padhta hai, jo yahan

0

hai, isliye woh actually decide karne wale quartic ko dekh nahi paata.

f^{''}

sign change karta hai" ek stronger requirement kyun hai "

f^{''} = 0

" se?

Ek function zero ko chhoo sakta hai bina cross kiye (jaise

12 x^{2}

0

). Sirf actual crossing hi concavity ko

\cup

\cap

flip karti hai, jo geometrically ek inflection hai.

Concave up guarantee kyun karta hai ki critical point minimum hai, bina neighbours test kiye?

Concave up matlab slope

f^{'}

increasing hai;

c

par yeh

0

se guzarta hai, isliye

f^{'} < 0

just left aur

f^{'} > 0

just right — downhill phir uphill = ek valley, turant decide ho jaata hai.

f^{''} = - sin x

humein kyun batata hai ki

sin x

π

par inflect karta hai lekin

0

2 π

par nahi (interior sense mein)?

- sin x

sign change karta hai jab

sin x

zero cross karta hai

π

se guzarte hue positive se negative ki taraf (interior), ek concavity flip produce karta hai; endpoints

0, 2 π

par koi two-sided interval nahi hai compare karne ke liye, isliye koi interior inflection claim nahi ki jaati.

Second derivative test sirf "local" kyun hai?

Taylor expansion ek approximation hai jo sirf

c

ke paas valid hai; yeh door ke behaviour ke baare mein kuch nahi kehta, isliye yeh ek local extremum certify karta hai, kabhi global nahi (uske liye poore domain mein Maxima and minima — optimization comparison chahiye).

Edge cases

Boundary aur degenerate scenarios jinhein test galat handle karne ka moqa deta hai.

f (x) = x^{3}

x = 0

: max hai, min hai, ya inflection?

Koi extremum nahi:

f^{'} (0) = 0

hai lekin

f^{''} = 6 x

0

ke across

- \to +

flip karta hai aur

f

continuous hai, isliye yeh horizontal tangent ke saath ek inflection point hai.

f (x) = ∣ x ∣

x = 0

: second derivative test kya kehta hai?

Apply nahi ho sakta —

f^{''} (0)

exist nahi karta (corner), isliye

0

"undefined-derivative" wala critical point hai. First-derivative test phir bhi minimum deta hai kyunki slope

- 1 \to + 1

jaata hai.

f (x) = x^{1/3}

x = 0

f^{''}

undefined hone ke bawajood inflection?

Haan.

f^{''} = - \frac{2}{9} x^{- 5/3}

0

par undefined hai lekin sign change karta hai (

x < 0

ke liye concave up,

x > 0

ke liye concave down) jabki

f

continuous rehta hai, isliye concavity flip hoti hai — ek valid inflection.

Constant function

f (x) = 5

: koi concavity ya inflection?

f^{''} \equiv 0

aur kahin koi sign change nahi, isliye yeh na strictly concave up hai na down aur koi inflection points nahi hain — flatness ek switch nahi hai.

f (x) = x^{2}

[1, 3]

: kya second derivative test iska minimum dhundhta hai?

Nahi — minimum endpoint

x = 1

par hai jahan

f^{'} \neq = 0

hai. Test sirf interior critical points inspect karta hai; endpoints ko direct comparison chahiye (Curve sketching / boundary check).

Agar

f^{''} > 0

pure

R

par ho (ek convex function), toh kitne inflection points ho sakte hain?

Zero. Constant-sign

f^{''}

matlab concavity kabhi flip nahi hoti, jo ek convex function ki defining feature hai — dekho Concavity and convex functions.

f^{''} (c) = 0

aur

f^{'''} (c) \neq = 0

c

par kya conclude kar sakte ho?

Nonzero odd-order term

f^{''}

c

ke across actually sign change karne par majboor karta hai, isliye (continuity ke saath)

c

ek inflection point hai — ek clean sufficient condition.

Connections

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