Curve sketching — systematic approach: Question bank

4.1.30 · D5 · Maths › Calculus I — Limits & Derivatives › Curve sketching — systematic approach

Shuru karne se pehle, ye cheezein apne dimaag mein pakki kar lo:

Domain = $x$ ki woh set jahan $f$ actually ek number deta hai (zero se divide nahi, negatives ka square root nahi, non-positive numbers ka log nahi).
$f^{'} (x)$ = tangent line ki slope = chadhne/utarne ki rate. Dekho First derivative and monotonicity.
$f^{''} (x)$ = slope ke change ki rate = bending. Dekho Second derivative test.
Asymptote = ek line jiske paas curve aata hai (vertical, horizontal, ya oblique). Dekho Rational functions and asymptotes.
Critical point = jahan $f^{'} (x) = 0$ ya $f^{'}$ undefined hai.

Poore page ke liye apne dimaag mein do pictures rakho. Pehli, $f^{'}$ aur $f^{''}$ ek real curve pe kaisi dikhti hain:

Figure — Curve sketching — systematic approach

Red tangent arrows dekho: jahan wo upar slope karte hain function badhta hai ( $f^{'} > 0$ ); jahan neeche girate hain woh ghatta hai ( $f^{'} < 0$ ); peak pe tangent flat hai ( $f^{'} = 0$ ). Bending — cup vs cap — woh $f^{''}$ ki kahani hai jo upar layer hoti hai.

Doosri, woh tool jo almost har "trap" abuse karta hai — sign chart. Yeh saari calculus ko $+$ aur $-$ marks ki ek number line mein compress karta hai:

Ek sign chart ka matlab sirf tab hota hai jab points domain ke andar hoon, aur woh sirf tab feature flag karta hai jab sign actually change hoti hai. Neeche ke aadhe traps log sign chart galat padhne ki wajah se hain — yeh picture apne paas rakho.

True or false — justify

Har answer mein reason dena zaroori hai, sirf verdict nahi.

Har woh point jahan

f^{'} (x) = 0

hai, local maximum ya minimum hota hai.

False.

f^{'}

ka wahaan sign change hona zaroori hai; jaise

f (x) = x^{3}

mein

f^{'} (0) = 0

hai phir bhi curve ek flat inflection se hote hue chadhti rehti hai.

Agar

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

ho toh

x_{0}

ek inflection point hai.

False.

f^{''}

x_{0}

par actually sign change hona zaroori hai; jaise

f (x) = x^{4}

mein

f^{''} (0) = 0

hai lekin dono sides pe concave-up rehta hai, toh koi inflection nahi.

Ek curve kabhi apne horizontal asymptote ko cross nahi kar sakti.

False. Horizontal asymptote sirf

x \to \pm \infty

ke waqt behaviour control karta hai;

f (x) = \frac{s i n x}{x}

settle hone se pehle

y = 0

ko infinitely baar cross karta hai.

Ek curve kabhi apne vertical asymptote ko cross nahi kar sakti.

True. Vertical asymptote ek aisi

x

-value pe hota hai jo domain se bahar hai, toh graph ka koi bhi point us

x

-coordinate pe nahi hota — cross karne ko kuch hai hi nahi.

Agar

f

even hai, toh phir bhi uska ek oblique (slant) asymptote ho sakta hai.

False. Even function

y

-axis ke baare mein symmetric hoti hai, toh agar woh right side pe kisi slant line

y = m x + c

ke paas jaaye toh left pe mirror line

y = - m x + c

bhi chahiye — ek single oblique asymptote dono satisfy nahi kar sakti, isliye evenness unhe rule out kar deti hai (sirf horizontal asymptotes survive karte hain).

Degree

\geq 1

ka ek polynomial hamesha kam se kam ek

x

-intercept rakhta hai.

False. Odd-degree polynomials zaroor cross karte hain (ends opposite directions mein jaate hain), lekin even-degree wale jaise

f (x) = x^{2} + 1

kabhi

x

-axis ko touch nahi karte.

Agar kisi interval par

f^{'} (x) > 0

ho, toh us mein kisi bhi

a < b

ke liye

f (b) > f (a)

hoga.

True. Bas

f

$[a, b]$ par continuous aur $(a, b)$ par differentiable honi chahiye, Mean Value Theorem deta hai

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

kisi

c \in (a, b)

ke liye;

f^{'} (c) > 0

aur

b - a > 0

ke saath difference positive hoga. Dekho Mean Value Theorem.

Agar

f

kisi interval par increasing hai, toh us poore interval mein

f^{'} (x) > 0

hoga.

False. Increasing hona sirf

f^{'} (x) \geq 0

force karta hai;

f (x) = x^{3}

strictly increasing hai phir bhi

f^{'} (0) = 0

hai.

Ek function jo har jagah concave up hai, uska apne domain ke interior mein koi local maximum nahi hoga.

True. Concave up matlab curve apni tangents ke upar rehti hai, toh koi bhi interior critical point minimum hoga. Caveat: maximum phir bhi ek endpoint ya boundary/limiting value par aa sakta hai — concavity sirf interior maxima ko rule out karti hai, boundary waalon ko nahi.

Agar ek rational function mein factor

(x - a)

sirf uske denominator mein hai (numerator se cancel nahi hua), toh

x = a

ek vertical asymptote hai.

True. Denominator-zero tabhi vertical asymptote deta hai jab factor cancel na ho; agar wahi factor numerator ko bhi divide kare toh woh cancel hokar ek removable hole ban jaata hai (edge-case section dekho), toh "uncancelled" condition hi asli rule hai.

Spot the error

Galat reasoning padho, phir bilkul sahi bolo ki kaun sa step toota.

f (x) = \frac{x ^{2}}{x ^{2} - 1}

ke liye asymptote

y = 1

hai, toh curve har jagah

y = 1

ke neeche rehni chahiye."

Galat. Horizontal asymptote sirf end behaviour hai. Walls ke beech (

∣ x ∣ < 1

) function actually negative hai; bahar (

∣ x ∣ > 1

) woh

y = 1

ke upar baithti hai. Line

y = 1

koi ceiling nahi hai.

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

zero hai

x = 0

par aur undefined hai

x = \pm 1

par, toh mere paas teen critical points hain:

0, 1, - 1

Galat. Critical points domain mein hone chahiye. Yahaan

x = \pm 1

excluded hain (asymptotes), toh woh critical points nahi hain — sirf

x = 0

hai.

f (x) = \frac{x ^{2}}{x ^{2} - 1}

ke liye

f^{''}

ka sign

x = \pm 1

par change hota hai, isliye

x = \pm 1

par inflection points hain."

Galat. Ek asli inflection point curve par ek point hona chahiye.

x = \pm 1

par function defined hi nahi hai, toh concavity ek gap ke across flip karti hai, actual point par nahi — wahaan koi inflection nahi hai.

f (x) = x^{3}

mein

f^{'} (0) = 0

aur

f^{''} (0) = 0

hai, aur kyunki

f^{''}

inconclusive hai toh main

x = 0

classify nahi kar sakta."

Second Derivative Test inconclusive hai, lekin First Derivative Test kaam khatam karta hai:

f^{'} = 3 x^{2} \geq 0

dono sides pe (koi sign change nahi), toh

x = 0

ek flat inflection hai, extremum nahi.

f (x) = \frac{x ^{2} + 1}{x - 1}

ka oblique asymptote dhundhne ke liye main

lim_{x \to \infty} \frac{f ( x )}{x} = 1

leta hoon, toh asymptote

y = x

hai."

Tumne

m = 1

sahi dhundha lekin jaldi rok diya. Tumhe abhi

c = lim_{x \to \infty} (f (x) - x)

bhi chahiye, jo

c = 1

deta hai, toh asymptote

y = x + 1

hai,

y = x

nahi.

"Maine domain, intercepts, aur derivatives compute kiye, lekin mujhe same peak mili chahe maine notice kiya ya nahi ki

x = 2

domain mein nahi hai — toh domain step optional hai."

Saamaanyatah galat. Ek excluded point bilkul waheen ho sakta hai jahan tum otherwise smooth curve draw karte, ek phantom feature bana ke. Domain step skip karna hi woh tarika hai jis se hidden holes aur asymptotes miss ho jaate hain — yeh hamesha step 1 hai.

Why questions

Ye check karte hain ki tum jaante ho kyun har tool sahi wala hai.

Increasing/decreasing decide karne ke liye hum

f^{'}

ki exact value ki jagah uske sign ka use kyun karte hain?

Kyunki monotonicity ko sirf change ki direction chahiye; Mean Value Theorem ke zariye

f^{'}

ka sign akela

f (b) - f (a)

ka sign fix karta hai, toh trend ke liye magnitude irrelevant hai.

f^{''}

(slope ki derivative) steepness ki jagah bending kyun measure karta hai?

Steepness khud

f^{'}

hai;

f^{''}

batata hai ki woh steepness badh rahi hai (slope upar ghoom raha hai, concave-up cup) ya ghatt rahi hai (concave-down cap), aur yahi "bending" ka matlab hai.

Checklist mein domain pehle kyun check kiya jaata hai?

Baad ke har step — intercepts, asymptotes, sign charts — sirf usi set par meaningful hain jahan

f

exist karta hai; ek excluded point ek hidden asymptote ya hole ho sakta hai jo poori picture badal de.

Polynomial long division oblique asymptote kyun reveal karta hai?

Division

f (x) = m x + c + \frac{R ( x )}{Q ( x )}

likhta hai jahan remainder fraction

x \to \infty

ke saath

\to 0

ho jaata hai, toh curve line

m x + c

par squeeze ho jaata hai — leftover term literally ends par gayab ho jaata hai.

Symmetry (

f (- x) = f (x)

- f (x)

) kaam kyun bachaati hai?

Tum sirf

x \geq 0

analyse karte ho, phir mirror karo: even functions

y

-axis ke across reflect hoti hain, odd functions origin ke baare mein

18 0^{\circ}

rotate hoti hain, toh aadha sign-chart automatically carry over hota hai.

lim_{x \to \infty} f (x)

jaisi limit kabhi vertical asymptote kyun produce nahi kar sakti?

Vertical asymptote tab hota hai jab

x

ek finite value ke paas jaaye jabki

f \to \pm \infty

; infinity par limit

x

vary karta hai,

y

nahi, toh woh sirf horizontal (ya oblique) end behaviour reveal kar sakta hai. Dekho Limits and continuity.

Woh critical point jahan

f^{'}

undefined hai, phir bhi check karne layak kyun hai?

Extrema sharp corners ya cusps par aa sakte hain (jaise

f (x) = ∣ x ∣

0

par) jahan slope exist hi nahi karta; unhe ignore karna real maxima aur minima miss kara dega.

Edge cases

Woh scenarios jinhe checklist quietly assume kar ke chhod deta hai.

f (x) = \frac{x ^{2} - 1}{x - 1}

mein

x = 1

par kya hota hai?

Numerator aur denominator dono vanish hote hain, toh

x = 1

ek removable hole hai, asymptote nahi:

f

simplify hokar

x + 1

ban jaata hai jisme

(1, 2)

par sirf ek point punched out hai.

Kya ek bounded interval par ek function ke infinitely many critical points ho sakte hain?

Haan.

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

(with

f (0) = 0

)

0

ke paas tezi se oscillate karta hai, jo origin par accumulate hote critical points deta hai — wahaan sign chart ek finite list nahi ho sakta.

Graph ek critical point par kya karta hai jahan

f^{'}

sign change

+ \to +

karta hai (koi sign change nahi) lekin

f^{'} = 0

hai?

Woh momentarily flat ho jaata hai aur phir chadhta rehta hai — ek stationary inflection (jaise

x^{3}

0

par), jo na max hai na min.

Agar

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

aur

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

dono hain, toh kya woh point automatically inflection hai?

Nahi. Dono ka zero hona inconclusive hai; tumhe sign changes test karne honge.

x^{4}

0

par

f^{'} = f^{''} = 0

hai phir bhi woh genuine minimum hai, jabki

x^{3}

0

par inflection hai.

Kya

+ \infty

par horizontal asymptote

- \infty

wale ke barabar hona zaroori hai?

Nahi. Ek function dono sides par alag horizontal levels approach kar sakta hai, jaise

f (x) = \frac{x}{x ^{2} + 1}

x \to + \infty

par

+ 1

aur

x \to - \infty

par

- 1

ki taraf jaata hai — do alag horizontal asymptotes.

Ek seedhi line ki concavity kya hai, aur kya uske inflection points hote hain?

Ek line mein

f^{''} \equiv 0

har jagah hota hai, toh woh na concave up hai na neeche aur uske koi inflection points nahi hote — concavity ke liye actual bending chahiye, jo ek line mein kabhi nahi hoti.

Kya ek function even aur odd dono ho sakta hai?

Sirf zero function

f (x) = 0

. Evenness deta hai

f (- x) = f (x)

aur oddness deta hai

f (- x) = - f (x)

; saath mein ye force karte hain

f (x) = - f (x)

, isliye saare

x

ke liye

f (x) = 0

Ek vertical asymptote par, kya curve dono sides se

+ \infty

ki taraf jaana zaroori hai?

Nahi. Dono one-sided limits ka sign alag ho sakta hai — jaise

f (x) = \frac{1}{x}

0

ke left se

- \infty

ki taraf jaata hai aur right se

+ \infty

ki taraf; tumhe har side alag check karni hogi.

Recall Trap pattern ka one-line summary

Yahaan almost har trap teen shapes mein se ek hai: (1) ek necessary condition ( $f^{'} = 0$ , $f^{''} = 0$ ) ko sufficient samajhna — hamesha ek sign change ki maang karo; (2) ek asymptote ko barrier samajhna instead of limiting behaviour ke; (3) domain bhool jaana, jo "asymptotes" ko holes aur phantom critical points mein badal deta hai. Reason karo, reflex mat karo.