Critical points — finding, classifying: Question bank

4.4.12 · D5 · Maths › Multivariable Calculus › Critical points — finding, classifying

Shuru karne se pehle, parent note se machinery yaad karo: ek critical point woh jagah hai jahan $\nabla f = 0$ (dono partials zero ho jaate hain) ya koi partial exist nahi karta; discriminant hai $D = f_{xx} f_{y y} - (f_{x y})^{2} = det H$ jahan $H$ Hessian Matrix hai; aur test yeh kehta hai: $D > 0, f_{xx} > 0 \Rightarrow$ min, $D > 0, f_{xx} < 0 \Rightarrow$ max, $D < 0 \Rightarrow$ saddle, $D = 0 \Rightarrow$ inconclusive.

Jo teen shapes aap classify kar rahe ho woh kuch aise dikhti hain — traps tackle karne se pehle in pictures ko memorise karo:

True or false — justify

TF1. "Agar

f_{xx} > 0

aur

f_{y y} > 0

kisi critical point par hain, toh woh zaroor local minimum hoga."

Jhooth. Aapne mixed term ignore ki.

f = x^{2} + y^{2} - 3 x y

mein

f_{xx} = f_{y y} = 2 > 0

hai lekin

D = 4 - 9 = - 5 < 0

— ek saddle. Cross-curvature dominate kar sakti hai.

TF2. "Agar

D < 0

ho toh point hamesha ek saddle hoga, chahe

f_{xx}

kuch bhi ho."

Sach.

D < 0

matlab quadratic form indefinite hai (kuch directions mein upar, kuch mein neeche), jo exactly saddle hai;

f_{xx}

yahan irrelevant hai.

TF3. "

D = 0

matlab point definitely max, min, ya saddle mein se kuch bhi nahi hai."

Jhooth.

D = 0

inconclusive hai, koi verdict nahi.

f = x^{4} + y^{4}

ka origin par

D = 0

hai phir bhi genuinely minimum hai; quadratic term bas decide nahi kar paai.

TF4. "Har woh point jahan

\nabla f = 0

ho, woh local max, min, ya saddle hota hai."

Jhooth. Yeh sirf candidates hain, lekin monkey saddle ya degenerate flat region (

D = 0

cases) teeno mein se clearly kuch nahi ho sakta;

\nabla f = 0

necessary hai, sufficient nahi.

TF5. "Second derivative test ko

f = x^{2} + y^{2}

ke corner par apply kiya ja sakta hai."

Jhooth. Test ko

f

ka twice differentiable hona chahiye. Cusp par partials exist hi nahi karte, isliye Hessian undefined hai — classify karne ke liye direct inspection karo.

TF6. "Agar

f_{xx} = 0

kisi critical point par ho, toh completing-the-square derivation toot jaati hai aur test bhi."

Jhooth. Derivation jaise likhi gayi usne

f_{xx} \neq = 0

assume kiya tha, lekin conclusion (

D = det H

aur uske sign rules) phir bhi valid rehta hai — aap

k

mein pehle square complete kar sakte ho, ya

D

seedha padh sakte ho.

TF7. "Ek function ka ek critical point ho sakta hai jo har straight-line direction mein local max ho lekin genuinely local max na ho."

Sach. Yeh classic Peano surface phenomenon hai: lines tak restrict karna mislead kar sakta hai kyunki surface ek curved path ke along neeche ja sakti hai. Straight-line tests kaafi nahi hain; Hessian (ya koi curve) chahiye.

TF8. "

f_{x y}

f_{y x}

se swap karna

D

ki value badal sakta hai."

Jhooth.

f

ke continuous second partials ke liye,

f_{x y} = f_{y x}

(Clairaut/Schwarz), isliye

H

symmetric hai aur

D

unambiguous hai.

TF9. "Agar Hessian ke dono eigenvalues positive hain, toh point minimum hai."

Sach. Positive eigenvalues matlab quadratic form positive-definite hai — har direction mein upar — jo

D > 0, f_{xx} > 0

ka hi dusra roop hai.

TF10. "

D > 0

akele 'extremum' conclude karne ke liye kaafi hai bina yeh jaane ki kaunsa."

Sach.

D > 0

force karta hai ki

f_{xx}

aur

f_{y y}

ek sign share karein (kyunki

f_{xx} f_{y y} = D + f_{x y}^{2} > 0

), isliye yeh definitely max ya min hai — bas

f_{xx}

ka sign chahiye yeh jaanne ke liye ki kaunsa.

Spot the error

SE1. "

f = x^{3} - 3 x + y^{2}

: maine

x = \pm 1

nikala, toh dono

(1, 0)

aur

(- 1, 0)

minima hain kyunki

f_{y y} = 2 > 0

Galti: classification mein sirf

f_{y y}

use kiya. Yahan

D = 12 x

hai, isliye

(1, 0)

par

D > 0

(min) hai lekin

(- 1, 0)

par

D < 0

(saddle) hai. Pehle hamesha

D

compute karo.

SE2. "

f = x^{2} - y^{2}

ke liye maine

D = (2) (- 2) = - 4

nikala, aur kyunki

f_{xx} = 2 > 0

hai, yeh minimum hai."

Galti: jab

D < 0

mile toh ruk jaao — yeh saddle hai.

D < 0

paane ke baad

f_{xx}

padhna meaningless hai; min/max branch sirf tab chalti hai jab

D > 0

ho.

SE3. "Origin par

f = x^{4} + y^{4}

D = 0

, isliye test inconclusive hai aur hum point ke baare mein kuch nahi keh sakte."

Galti: inconclusive matlab test decide nahi kar sakta, yeh nahi ki hum nahi kar sakte. Direct inspection (

f \geq 0

, sirf origin par equal) prove karta hai ki yeh global minimum hai.

SE4. "Maine

f_{x} = 0

solve karke

x = 2

nikala aur alag se

f_{y} = 0

y = 3

, toh critical point

(2, 3)

hai."

Galti: equations ko simultaneously solve karna hoga. Agar

f_{x} = 0

ke solutions

y

par depend karte hain, toh aap

x

aur

y

independently set nahi kar sakte; ek ko dusre mein substitute karo.

SE5. "

f = e^{x}

: maine

f_{x} = e^{x} = 0

set kiya; koi solution nahi, toh koi critical point nahi aur test apply nahi hota — function kahin flat hogi."

Galti: yeh sahi hai ki

f_{x} = e^{x} > 0

kabhi zero nahi hota (toh koi critical point nahi), lekin "kahin flat hogi" ka conclusion galat hai. Koi critical point nahi matlab koi extrema nahi hain; function bas badhta rehta hai.

SE6. "Hessian

H = (2312)

D = det H = 4 - 3 = 1 > 0

milta hai."

Galti: Hessian symmetric hona chahiye (

f_{x y} = f_{y x}

), isliye off-diagonals

1

aur

3

nahi ho sakte. Agar genuinely

f_{x y} = f_{y x} = c

ho, toh

D = 4 - c^{2}

hoga.

SE7. "Kyunki extrema ke liye

\nabla f = 0

required hai,

\nabla f

ka sign check karna max vs min batayega."

Galti: critical point par

\nabla f

zero hota hai, isliye padhne ke liye koi sign hai hi nahi. Classification second derivatives ka kaam hai, gradient ka nahi.

Kyun wale sawaal

W1. "Kyun first derivatives critical point ka type ke baare mein kuch nahi batate?"

Kyunki woh sab wahan zero hain — flat tangent plane maxes, mins, aur saddles sab mein common hai. Sirf curvature (second derivatives) shapes ko distinguish karti hai.

W2. "Mixed partial

f_{x y}

D

mein aata hi kyun hai?"

Yeh measure karta hai ki

x

mein curvature

y

se kaise couple hoti hai — bowl ki ek tilting. Badi tilting ek would-be bowl ko saddle mein rotate kar sakti hai, jo exactly

- (f_{x y})^{2}

term hai jo

D

ko neeche kheenchti hai.

W3. "Kyun

D = det H

hai na ki, maan lo,

H

ka trace?"

Determinant eigenvalues ka product hai, isliye uska sign detect karta hai ki woh sign share karte hain (definite → extremum) ya differ karte hain (indefinite → saddle). Trace (unka sum) ek saddle ko extremum se alag nahi kar sakta. Neeche ki figure eigenvectors ko surface ke principal axes ke roop mein aur unke curvature signs ke

D

mein multiply hone ko dikhati hai.

W4. "Taylor derivation mein hum sirf

f_{xx}

aur

f_{y y}

dekhne ki bajaye square kyun complete karte hain?"

Square complete karna quadratic ko independent squares ke sum ke roop mein rewrite karta hai, jo form ka sign har direction mein ek saath expose karta hai — cross term ek square mein fold ho jaata hai, aur

D / f_{xx}

honest second coefficient ke roop mein reh jaata hai. Algebraically nikala gaya aur neeche picture hai:

$f_{xx} h^{2} + 2 f_{x y} hk + f_{y y} k^{2} = f_{xx} (h + \frac{f _{x y}}{f _{xx}} k)^{2} + = D / f_{xx} (f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}) k^{2} .$

W5. "Kyun smooth

f

ke extrema sirf wahan hote hain jahan

\nabla f = 0

ho?"

Agar koi bhi directional slope nonzero hoti toh aap thoda step leke neeche (ya upar) ja sakte the, jo point ke local extreme hone ko contradict karta. Isliye har partial vanish hona chahiye — Fermat's theorem several variables mein.

W6. "Kyun 1-D test mein

D

jaisi koi cheez mention nahi hoti?"

Ek dimension mein sirf ek direction mein curve karne ki possibility hai, isliye

f^{''}

ka sign poori kahani hai. 2-D

D

exactly isliye exist karta hai kyunki do directions disagree kar sakte hain (

x

mein upar,

y

mein neeche) — ek aisi possibility jo ek line par simply arise nahi ho sakti.

W7. "Kyun test ko Hessian ke eigenvalues use karke recast kiya ja sakta hai?"

Quadratic form

\frac{1}{2} v^{⊤} H v

positive-definite hai iff dono eigenvalues

> 0

(min), negative-definite iff dono

< 0

(max), indefinite iff woh sign mein differ karein (saddle) — aur

D = det H = λ_{1} λ_{2}

exactly yeh sign agreement encode karta hai.

Edge Cases

E1. "Kya hoga agar

D > 0

ho lekin

f_{xx} = 0

— kya yeh ho sakta hai?"

Nahi.

D > 0

matlab

f_{xx} f_{y y} = D + (f_{x y})^{2} > 0

, isliye

f_{xx}

zero nahi ho sakta. Jab

D > 0

ho tab '

f_{xx}

padho' step hamesha well-defined hota hai.

E2. "

f = x^{3} + y^{3}

ke origin par, kya yeh max, min, ya saddle hai?"

Wahan sab second partials vanish hote hain, isliye

D = 0

— inconclusive. Direct inspection: line

y = x

ke along,

f = 2 x^{3}

sign change karta hai

0

ke through, isliye yeh na max hai na min (ek degenerate saddle-like point). Neeche ka contour plot origin ke around sign-change lobes dikhata hai.

E3. "Monkey saddle

f = x^{3} - 3 x y^{2}

ke origin par test kya kehta hai?"

f_{xx} = 6 x

f_{y y} = - 6 x

f_{x y} = - 6 y

sab origin par vanish hote hain,

D = 0

dete hain — inconclusive. Yeh actually teen descending valleys wala saddle hai, lekin quadratic test ise dekh nahi sakta; cubic terms decide karte hain. Yeh teen-fold pattern figure mein unmistakable hai:

E4. "Kya critical point wahan exist ho sakta hai jahan

f

differentiable nahi, isliye Hessian exist nahi karta?"

Haan — definition explicitly un points ko include karti hai jahan koi partial exist karna fail kare (corners, cusps jaise

x^{2} + y^{2}

0

). Aise points candidates hain lekin

D

se nahi, definition se classify karne chahiye.

E5. "Agar

f_{xx} < 0

aur

f_{y y} < 0

ho lekin

D < 0

ho, toh kya hai?"

Saddle. Jab

D < 0

ho tab pure curvatures ke shared negative sign ka koi matlab nahi — cross-term ne form ko indefinite bana diya hai.

E6. "Kya constant function

f = 5

ke sab points critical hain, aur kaunse type ke?"

Har point par

\nabla f = 0

hai, isliye haan — sab critical hain. Lekin

H

zero matrix hai,

D = 0

har jagah: test se inconclusive, aur inspection se har point simultaneously (weak) max aur min hai kyunki

f

kabhi change hi nahi hota.

E7. "Do Hessian eigenvalues ke liye

D < 0

geometrically kya matlab hai?"

Unke opposite signs hain (

λ_{1} λ_{2} = D < 0

), isliye surface ek principal direction ke along upar curve karti hai aur perpendicular wali direction mein neeche — saddle ki defining picture.

Active Recall

Poora test ek sign chart mein compress hota hai — is picture ko memory mein commit karo:

Recall Do cheezein jo

$D$ positive hone par kabhi nahi kar sakta. Yeh $f_{xx}$ ko ambiguous nahi chhod sakta (kyunki $f_{xx} \neq = 0$ ), aur yeh saddle indicate nahi kar sakta. $D > 0$ hamesha ek genuine extremum matlab hai.

Recall "Positive

$f_{xx}$ aur $f_{y y}$ ⟹ minimum" disprove karne ka sabse fast tarika. $f = x^{2} + y^{2} - 3 x y$ cite karo: dono pure curvatures positive, phir bhi $D = 4 - 9 = - 5 < 0$ , ek saddle.

Recall Kyun

$D = 0$ ek classification nahi balki ek confession hai? Yeh confess karta hai ki quadratic approximation decide karne ke liye bahut flat hai; aapko higher-order terms ya direct inspection par fall back karna hoga.

Connections

Critical points — finding, classifying — woh parent note jise yeh bank drill karta hai.
Hessian Matrix — $D = det H$ ; upar har trap ka eigenvalue view.
Quadratic Forms and Definiteness — definite/indefinite ↔ max/min/saddle verdicts.
Taylor Series in Several Variables — completing-the-square argument ka source.
Gradient and Directional Derivatives — kyun $\nabla f = 0$ necessary hai.
Single-variable Second Derivative Test — 1-D case bina $D$ ke.
Lagrange Multipliers — constrained-extremum cousin.