Applications — increasing - decreasing, local extrema (first derivative test): Question bank

4.1.28 · D5 · Maths › Calculus I — Limits & Derivatives › Applications — increasing - decreasing, local extrema (first

True or false — justify

Agar

f^{'} (c) = 0

ho toh

c

ek local maximum ya minimum hai.

False. Zero slope ka matlab sirf "yahan flat hai"; extremum banane ke liye slope ka sign switch karna zaroori hai.

f (x) = x^{3}

mein

f^{'} (0) = 0

hai phir bhi function badhta rehta hai — yeh ek inflection point hai jahan tangent horizontal hai.

Agar kisi interval

I

par

f^{'} (x) > 0

ho, toh

f

I

par increasing hai.

True. Mean Value Theorem ke zariye,

f (b) - f (a) = f^{'} (c) (b - a)

jahan

b - a > 0

, toh positive

f^{'}

har

a < b

ke liye

f (b) > f (a)

force karta hai.

Agar

f

I

par increasing hai, toh har

x \in I

ke liye

f^{'} (x) > 0

hoga.

False — sirf

f^{'} (x) \geq 0

guarantee hai.

f (x) = x^{3}

strictly increasing hai

R

par lekin

f^{'} (0) = 0

hai: ek akela flat instant overall rising ko nahi rokta.

Har local maximum wahan hota hai jahan

f^{'} = 0

ho.

False. Extrema critical points par hote hain, jisme woh jagahein bhi shamil hain jahan

f^{'}

undefined ho.

f (x) = ∣ x ∣

ka minimum

0

par hai jahan derivative exist hi nahi karta.

Har critical point ek extremum hota hai.

False.

f^{'} = 0

ya undefined hona sirf ek candidate hai.

x^{3}

(slope zero, no sign change) aur

x^{1/3}

(vertical tangent, no sign change) critical hain lekin flat-through hain, extrema nahi.

Agar

c

par

f^{'}

-

+

ho jaaye, toh

c

ek local minimum hai.

True — pehle girna phir chadhna matlab valley mein sabse neeche aaye. Yeh exactly First Derivative Test ka

- \to +

case hai.

Agar

f^{'} (c)

exist na kare, toh

c

extremum nahi ho sakta.

False.

∣ x ∣

jaisa corner

0

par ek genuine minimum hai precisely isliye kyunki slope kink ke across sign flip karta hai, chahe woh point par undefined ho.

Mean Value Theorem ko

f

ki differentiability closed interval

[a, b]

par chahiye.

False. Isko

[a, b]

par continuity chahiye lekin differentiability sirf open interval

(a, b)

par — endpoints par corners ya vertical tangents ho sakte hain.

Agar

c

ke bilkul left mein

f^{'} > 0

aur bilkul right mein bhi

f^{'} > 0

ho, toh

c

extremum nahi hai.

True. Dono sides mein same sign matlab function

c

se hota hua badhta raha; koi peak ya valley bana hi nahi — yeh "same hai, toh kuch nahi" wali baat hai.

Spot the error

f^{'} (x) = 0

solve karo, critical points milenge — kaam khatam."

Woh points miss ho gaye jahan

f^{'}

undefined hai. Cusps, corners, aur vertical tangents (

x^{2/3}

∣ x ∣

) bhi critical hain aur extrema rakh sakte hain.

f^{'} (2) = 0

aaya, aur jab

f^{'} (2)

test kiya toh zero nikla, toh

x = 2

par koi extremum nahi hai."

Tumne critical point khud test kiya, jo hamesha no sign deta hai. Tumhe

c

ke strictly left aur right mein sample karna hoga, kabhi

c

par nahi.

f^{'}

ka sign

+ \to -

ho gaya

x = 3

par, toh maximum

3

hai."

Maximum ki location

x = 3

hai; maximum value

f (3)

hai.

x

-coordinate ko "the maximum" kehna input aur output ko confuse karna hai.

f (x) = 1/ x

decreasing hai

(- \infty, 0) \cup (0, \infty)

par kyunki

f^{'} = - 1/ x^{2} < 0

har jagah hai."

Domain mein break ke across intervals ko join nahi kar sakte.

f

har piece par separately decreasing hai, lekin

f (- 1) = - 1 < f (1) = 1

, toh yeh union par decreasing nahi hai.

f^{'} > 0

har jagah hai sivaaye ek point ke jahan

f^{'} = 0

hai, toh

f

strictly increasing nahi hai."

Ek akela flat instant strict increase ko nahi rokta.

x^{3}

mein

f^{'} (0) = 0

hai phir bhi

a < b \Rightarrow a^{3} < b^{3}

hamesha — abhi bhi strictly increasing hai.

x = 5

par ek critical point mila;

f^{'}

ka sign left mein

-

aur right mein

+

tha, toh yeh maximum hai."

- \to +

ek minimum hai (valley mein neeche, phir upar).

+ \to -

maximum hai. Mnemonic: PND peak, NPD valley.

"Kyunki MVT ek

c

deta hai, toh function mein exactly ek point hai jo average slope match karta hai."

MVT kam se kam ek aisa

c

guarantee karta hai, exactly ek nahi. Ek wavy graph kaafi interior points par average slope match kar sakta hai.

Why questions

f^{'}

ka sign kyun decide karta hai increasing vs decreasing, uski size kyun nahi?

f (b) - f (a) = f^{'} (c) (b - a)

mein factor

(b - a) > 0

hai, toh sirf

f^{'} (c)

ka sign change ka sign flip kar sakta hai. Magnitude batata hai kitni tezi se, yeh nahi ki kis direction mein.

Critical points ke beech ke poore interval ki sign fix karne ke liye ek test point kyun kaafi hai?

Us interval par

f

differentiable hai, toh

f^{'}

exist karta hai aur kabhi zero nahi hota (saare zeros critical points hain jo hum pehle hi hata chuke hain). Agar

f^{'}

wahan bhi continuous ho, toh Intermediate Value Theorem usse zero se guzre bina sign change karne se rokta hai — toh woh ek sign rakhta hai aur ek sample interval settle kar deta hai.

f^{'}

critical points ke beech continuous hai" kab guaranteed hota hai, taaki one-test-point rule valid ho?

Yeh tab guaranteed hota hai jab

f

twice differentiable ho (ya

f^{'}

kisi aur tarah se continuous ho), jo polynomials, rationals, aur standard functions ko cover karta hai. Pathological

f^{'}

ke liye jo exist karta ho lekin discontinuous ho, ek point par trust karne ke bajaye sign zyada carefully check karna padega.

First Derivative Test ke liye

c

par

f

continuous kyun hona zaroori hai?

Test "

c

ke around rise then fall" padhta hai; agar

f

wahan jump kare, toh pieces ek genuine peak ya valley mein connect nahi hote, toh

f^{'}

ka sign akele mislead kar sakta hai.

Fermat's theorem sirf kyun kehta hai ki extrema critical points par hote hain, yeh nahi ki critical points hain extrema?

Yeh ek one-way filter hai: yeh candidates ko wahin narrow karta hai jahan slope zero ya undefined ho, lekin har candidate confirm karne ke liye sign change (ya Second Derivative Test) abhi bhi chahiye.

Sirf endpoints check karke

f (a) < f (b)

dekh lene se "increasing" conclude kyun nahi kar sakte?

Increasing ka matlab hai

I

mein saare

a < b

ke liye

f (a) < f (b)

. Do chosen points ke beech ek dip ek comparison satisfy karte hue poori definition violate kar sakta hai.

x^{3}

logon ko

0

ko minimum kyun samjha deta hai?

Kyunki

f^{'} (0) = 0

hai aur graph wahan flat ho jaata hai, valley ka bottom mimic karta hai. Lekin

f^{'} = 3 x^{2} > 0

dono sides par hai — koi sign change nahi, toh yeh inflection hai, min nahi.

Edge cases

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

ke liye

x = 0

par

f^{'}

ka sign behavior kya hai, aur test kya kehta hai?

f^{'} = \frac{1}{3} x^{- 2/3} \to + \infty

0

ke paas hai aur dono sides par positive hai (vertical tangent). Same sign

+ \to +

→ extremum nahi, chahe

f^{'}

0

par undefined hi kyun na ho.

Kya

f (x) = x^{2/3}

mein

x = 0

ek extremum hai?

Haan, ek local (aur global) minimum.

f^{'} = \frac{2}{3} x^{- 1/3}

x < 0

ke liye negative aur

x > 0

ke liye positive hai: ek

- \to +

cusp, toh yeh valley bottom hai.

Kya koi function

I

par increasing ho sakta hai phir bhi interior points par

f^{'} (c) = 0

ho?

Haan.

f (x) = x^{3}

strictly increasing hai aur

f^{'} (0) = 0

hai; isolated flat points allowed hain jab tak

f^{'} \geq 0

ho aur

f^{'}

kisi poore subinterval par zero na ho.

Domain ke endpoint par, jaise

x = 0

for

f

[0, 1]

, First Derivative Test ka kya hota hai?

Test karne ke liye sirf ek side hoti hai, toh sign-change rule apply nahi hota. Endpoints ko closed-interval method se handle kiya jaata hai, endpoint values directly compare karke.

Agar

f^{'}

har jagah positive ho sivaaye ek point ke jahan undefined ho, lekin

f

wahan continuous ho aur us se hoti hui upar jaati ho, toh woh point kya hai?

Ek critical point jo extremum nahi hai (jaise

x^{1/3}

0

par) — ek vertical-tangent inflection jahan rising

y

-direction mein kabhi rukti nahi.

Kya kisi bounded interval mein function ke infinitely many critical points ho sakte hain?

Haan.

f (x) = x^{2} sin (1/ x)

(

f (0) = 0

ke saath) itna oscillate karta hai ki uska slope

0

ke paas infinitely baar zero ho jaata hai — critical points accumulate ho sakte hain, toh "saare critical points dhundho" hamesha finite nahi hota.

Kya constant function

f (x) = 5

local max ya min count hoti hai?

Haan — usual (non-strict) definition ke under, har point simultaneously ek local maximum aur ek local minimum hai, kyunki

f (x) \leq f (c)

aur

f (x) \geq f (c)

kisi bhi neighbourhood par hold karte hain. Yeh na strictly increasing hai na decreasing.

Jis interval par

f^{'} \equiv 0

ho (ek genuine plateau, sirf ek point nahi), wahan

f

kya kar raha hai?

f

us poore interval par constant hai (kisi bhi sub-pair par MVT apply karke). Toh graph mein literally ek flat shelf hai, ek isolated flat instant nahi.

Agar koi function pehle utha, phir plateau par flat chala, phir gira (mesa ki tarah), toh kya plateau ek maximum hai?

Haan — us flat top ka har point ek local maximum hai (ek "flat maximum"). Test abhi bhi kaam karta hai: shelf se pehle

f^{'}

+

jaata hai aur baad mein

-

, poore top par

f^{'} = 0

ke saath. Mirror-image plateau ek flat minimum deta hai.

Kya plateau kabhi extremum nahi ho sakta?

Haan — agar flat shelf se pehle

f^{'}

+

ho aur baad mein bhi

+

ho (rising staircase par ek flat step), toh plateau na max hai na min; function bas aage badhne se pehle ruka tha.

Connections

Fermat's Theorem on Stationary Points — kyun critical points sirf candidates hain.
Mean Value Theorem — upar har sign-of- $f^{'}$ claim ke peeche ka engine.
Second Derivative Test — ek alternative jo smooth points ke liye "same sign" ambiguity se bachata hai.
Concavity and Inflection Points — $x^{3}$ aur $x^{1/3}$ "flat-through" cases explain karta hai.
Optimization (Closed Interval Method) — endpoints (upar exclude kiye gaye) kaise handle hote hain.
Curve Sketching — jahan ye saare traps ek saath dikhayi dete hain.