Curve sketching — systematic approach: Worked examples

4.1.30 · D3 · Maths › Calculus I — Limits & Derivatives › Curve sketching — systematic approach

Ye page parent ke tools assume karta hai: First derivative and monotonicity, Second derivative test, Rational functions and asymptotes, Limits and continuity, aur Optimization. Jahan use hote hain, wahan link diya hai.

Scenario matrix

Har curve-sketching problem inme se kuch case classes ka combination hoti hai. Har row ek alag behaviour hai jo tumhe handle karna aana chahiye; har ek ko neeche kam se kam ek worked example se cover kiya gaya hai.

#	Case class	Ismein kya khaas hai	Covered by
A	Polynomial, do turning points	$f^{'}$ ka sign do baar badalti hai; ends $\pm \infty$ ki taraf jaate hain	Ex 1
B	Degenerate critical point (flat inflection)	$f^{'} = 0$ lekin koi extremum nahi	Ex 2
C	Rational, horizontal asymptote	denominator degree $\geq$ numerator degree	Ex 3
D	Rational, vertical asymptotes dono signs ke saath	$f \to + \infty$ ek taraf, $- \infty$ doosri taraf	Ex 3
E	Rational, oblique/slant asymptote	numerator degree $=$ denominator degree $+ 1$	Ex 4
F	Even vs odd symmetry	reflection se kaam adha ho jaata hai	Ex 1 (odd), Ex 3 (even)
G	Apni horizontal asymptote ko cross karta hai	asymptote sirf infinity par govern karta hai, har jagah nahi	Ex 5
H	Zero / degenerate input (limit form $0/0$ )	removable hole vs genuine blow-up	Ex 6
I	Real-world word problem (optimization flavour)	curve shape ek physical question ka jawab deta hai	Ex 7
J	Exam twist — function mein parameter	behaviour ek constant ke sign par depend karta hai	Ex 8

Ab hum har example ko walk karenge, cells ke labels ke saath.

Example 1 — Polynomial, do turning points, odd symmetry · cells A, F

Step 1 — Domain & symmetry. $f$ ek polynomial hai, isliye yeh sabhi real $x$ ke liye defined hai. Symmetry check karo: $f (- x) = (- x)^{3} - 3 (- x) = - x^{3} + 3 x = - (x^{3} - 3 x) = - f (x) .$ Toh $f$ odd hai — origin ke baare mein symmetric hai. Ye step kyun? Odd symmetry ka matlab hai ek baar right half pata chal gayi, toh left half uski $18 0^{\circ}$ rotation hai — First derivative and monotonicity ka kaam adha ho jaata hai.

Step 2 — Intercepts. $f (0) = 0$ ; aur $x^{3} - 3 x = x (x^{2} - 3) = 0$ se milta hai $x = 0, \pm 3$ . Ye step kyun? Zeros anchor karte hain ki S-curve axis ko kahan cross karti hai.

Step 3 — First derivative. $f^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) .$ $f^{'} = 0$ set karo: $x = \pm 1$ . $f^{'}$ ka sign: $[- 1, 1]$ ke bahar $x^{2} > 1$ toh $f^{'} > 0$ (rising); andar, $f^{'} < 0$ (falling). Toh $x = - 1$ par $+ \to -$ change → local max; $x = 1$ par $- \to +$ → local min. Heights: $f (- 1) = (- 1) - 3 (- 1) = 2$ , aur oddness se $f (1) = - 2$ . Ye step kyun? $f^{'}$ ka sign monotonicity engine hai; uske zeros turning candidates hain.

Step 4 — Second derivative. $f^{''} (x) = 6 x .$ $x < 0$ ke liye $f^{''} < 0$ (concave down, frown), $x > 0$ ke liye $f^{''} > 0$ (concave up, smile), $x = 0$ par switch → inflection at $(0, 0)$ . Ye step kyun? Second derivative test max/min confirm karta hai ( $f^{''} (- 1) = - 6 < 0$ ✓ max, $f^{''} (1) = 6 > 0$ ✓ min) aur bend switch locate karta hai.

Neeche ka figure — kya observe karna hai. Burnt-orange S-curve lower left se upar climb karti hai, teal peak at $(- 1, 2)$ par pahunchti hai, ink inflection square at the origin se girte hue jahan woh briefly seedhi ho jaati hai, plum valley $(1, - 2)$ par neeche aati hai, phir dobara climb karti hai. Dekho ki picture origin ke baare mein $18 0^{\circ}$ rotation hai — yahi oddness visible hoti hai.

Figure — Curve sketching — systematic approach

Figure 1 — Cubic $f (x) = x^{3} - 3 x$ : local max $(- 1, 2)$ , local min $(1, - 2)$ , inflection at the origin.

Verify: directly compute karo: $f (- 1) = (- 1)^{3} - 3 (- 1) = - 1 + 3 = 2$ ✓; $f (1) = 1 - 3 = - 2$ ✓; $f^{''} (- 1) = 6 (- 1) = - 6 < 0$ (max) ✓; $f^{''} (1) = 6 > 0$ (min) ✓; $f (0) = 0$ (inflection height) ✓. (Machine-checked below.)

Example 2 — Degenerate critical point (flat inflection) · cell B

Step 1 — First derivative aur uska sign. $f^{'} (x) = 3 x^{2} .$ Yeh har jagah $\geq 0$ hai, sirf $x = 0$ par $0$ ke barabar. Toh $f^{'}$ ka sign $x = 0$ par nahi badalti — dono taraf $+$ hai. Ye step kyun? First Derivative Test kehta hai ki extremum ke liye sign change chahiye. Change nahi → max nahi, min nahi.

Step 2 — Second derivative. $f^{''} (x) = 6 x,$ jo $x = 0$ par $- \to +$ sign change karta hai. Toh $x = 0$ ek inflection point hai — lekin horizontal tangent ke saath. Yeh ek flat (stationary) inflection hai. Ye step kyun? Yeh parent ke mistake-callout ko action mein dikhata hai: $f^{'} = 0$ akela guarantee nahi karta extremum.

Verify: $f^{'} (- 1) = 3 (- 1)^{2} = 3 > 0$ aur $f^{'} (1) = 3 > 0$ (dono taraf same sign ⇒ koi extremum nahi); $f^{''} (- 1) = - 6 < 0$ aur $f^{''} (1) = 6 > 0$ (concavity flip ⇒ inflection). (Machine-checked below.)

Example 3 — Rational with horizontal wall & twin vertical walls, even · cells C, D, F

Step 1 — Domain & symmetry. $x^{2} - 1 = 0$ , $x = \pm 1$ par, toh domain hai $R ∖ {- 1, 1}$ . Aur $f (- x) = \frac{x ^{2}}{x ^{2} - 1} = f (x)$ → even, $y$ -axis ke upar mirror. Ye step kyun? Even symmetry se hum $x \geq 0$ study karke reflect kar sakte hain.

Step 2 — Asymptotes (limits explicitly). Vertical: jaise $x \to 1^{-}$ , denominator $\to 0^{-}$ , numerator $\to 1$ , toh $f \to - \infty$ ; jaise $x \to 1^{+}$ , denominator $\to 0^{+}$ , toh $f \to + \infty$ . (Evenness se $- 1$ ke walls yahi mirror karte hain.) Horizontal: upar aur neeche $x^{2}$ se divide karo: $lim_{x \to \pm \infty} \frac{x ^{2}}{x ^{2} - 1} = lim \frac{1}{1 - 1/ x ^{2}} = 1, toh y = 1.$ Ye step kyun? Asymptotes woh skeleton hain jis par curve drapeti hai; sided limits batate hain ki curve kis direction mein har wall se escape karta hai.

Step 3 — First derivative (quotient rule). $f^{'} (x) = \frac{2 x ( x ^{2} - 1 ) - x ^{2} ( 2 x )}{( x ^{2} - 1 ) ^{2}} = \frac{- 2 x}{( x ^{2} - 1 ) ^{2}} .$ Denominator ek square hai, hamesha $> 0$ . Toh $f^{'}$ ka sign $- 2 x$ ka sign hai: $x < 0$ ke liye positive, $x > 0$ ke liye negative → local max at $(0, 0)$ . Ye step kyun? $f^{'}$ ka sign climb/fall deta hai; domain mein sirf ek critical point hai $x = 0$ .

Step 4 — Value regions. $∣ x ∣ < 1$ ke liye: $x^{2} - 1 < 0$ toh $f < 0$ (curve $x$ -axis ke neeche baithta hai, origin par peak karta hai). $∣ x ∣ > 1$ ke liye: $f > 1$ (ceiling ke upar). Second derivative $f^{''} (x) = \frac{6 x ^{2} + 2}{( x ^{2} - 1 ) ^{3}}$ , $∣ x ∣ > 1$ par $> 0$ (concave up) aur $∣ x ∣ < 1$ par $< 0$ (concave down); koi inflection nahi kyunki sign sirf forbidden $x = \pm 1$ ke across flip hota hai. Ye step kyun? Curve ko har region mein uske asymptotes ke relative locate karta hai.

Neeche ka figure — kya observe karna hai. $x = \pm 1$ par do plum dashed verticals walls hain; teal dashed horizontal $y = 1$ par ceiling hai. Beech ki strip mein orange curve ek low hump banata hai jo ink dot $(0, 0)$ par peak karta hai aur har wall ke paas jaate hi $- \infty$ mein dive karta hai. Walls ke bahar dono orange branches $y = 1$ ke upar hain aur door jaate jaate uski taraf sweep karte hain. Poori picture $y$ -axis ke across mirror-symmetric hai — yahi evenness hai.

Figure 2 — $f (x) = x^{2} / (x^{2} - 1)$ : walls at $x = \pm 1$ , ceiling $y = 1$ , even hump peaking at the origin.

Verify: $f (0) = 0$ ; $f^{'} (0) = \frac{- 2 \cdot 0}{( 0 - 1 ) ^{2}} = 0$ ; $f^{''} (0) = \frac{6 \cdot 0 + 2}{( 0 - 1 ) ^{3}} = \frac{2}{- 1} = - 2 < 0$ (concave down ⇒ max); $lim_{x \to \infty} f = 1$ ; $lim_{x \to 1^{+}} f = + \infty$ . (Machine-checked below.)

Example 4 — Rational with ek oblique (slant) asymptote · cell E

Step 1 — Long division se line reveal karo. $\frac{x ^{2} + 1}{x} = x + \frac{1}{x} .$ Jaise $x \to \pm \infty$ , term $\frac{1}{x} \to 0$ , toh curve line $y = x$ se chipak jaata hai. Woh line oblique asymptote hai. Ye step kyun? Polynomial division "line part" ko ek remainder se alag karta hai jo infinity par vanish ho jaata hai — bilkul parent ka oblique-asymptote recipe. Formula se cross-check karo, kis direction of infinity har limit use karta hai yeh carefully state karte hue: $m = lim_{x \to + \infty} \frac{f ( x )}{x} = lim_{x \to + \infty} \frac{x ^{2} + 1}{x ^{2}} = 1, c = lim_{x \to + \infty} (f (x) - x) = lim_{x \to + \infty} \frac{1}{x} = 0.$ Yahi do limits $x \to - \infty$ par lene par identical $m = 1, c = 0$ milta hai, toh single line $y = x$ dono ends par asymptote hai.

Step 2 — Vertical asymptote & domain. Domain $R ∖ {0}$ . Jaise $x \to 0^{+}$ , $f = x + \frac{1}{x} \to + \infty$ ; jaise $x \to 0^{-}$ , $f \to - \infty$ . Toh $x = 0$ ek two-signed vertical wall hai. Ye step kyun? Denominator zero with non-zero numerator ⇒ genuine blow-up.

Step 3 — Symmetry & first derivative (full monotonicity). $f (- x) = - x - \frac{1}{x} = - f (x)$ → odd. $f^{'} (x) = 1 - \frac{1}{x ^{2}} = \frac{x ^{2} - 1}{x ^{2}} .$ Denominator $x^{2} > 0$ hamesha, toh $f^{'}$ ka sign $x^{2} - 1$ ka sign hai. Saare char intervals walk karo, critical points $x = \pm 1$ aur wall $x = 0$ se split karke:

$(- \infty, - 1)$ : $x^{2} > 1$ ⇒ $f^{'} > 0$ → increasing.
$(- 1, 0)$ : $x^{2} < 1$ ⇒ $f^{'} < 0$ → decreasing.
$(0, 1)$ : $x^{2} < 1$ ⇒ $f^{'} < 0$ → decreasing.
$(1, \infty)$ : $x^{2} > 1$ ⇒ $f^{'} > 0$ → increasing.

Toh left branch par curve $x = - 1$ tak rise karke wall mein fall karta hai, local max at $(- 1, - 2)$ deta hai; right branch par wall se $x = 1$ tak fall karta hai phir rise karta hai, local min at $(1, 2)$ deta hai. Ye step kyun? $f^{'}$ ka interval-by-interval sign poori monotonicity picture hai, aur $+ \to -$ / $- \to +$ changes extrema pin karte hain.

Step 4 — Second derivative (concavity, all cases). $f^{''} (x) = \frac{2}{x ^{3}} .$ $f^{''}$ ka sign sirf $x^{3}$ ka sign hai:

$x > 0$ : $f^{''} > 0$ → concave up (right branch upward cup karta hai, slant line $y = x$ ke upar baitha hai).
$x < 0$ : $f^{''} < 0$ → concave down (left branch downward cap karta hai, $y = x$ ke neeche baitha hai).

Koi inflection point nahi hai: $f^{''}$ sirf forbidden $x = 0$ ke across sign change karta hai, domain ke kisi point par nahi. Concavity extrema bhi confirm karta hai — $f^{''} (1) = 2 > 0$ (min ✓), $f^{''} (- 1) = - 2 < 0$ (max ✓). Ye step kyun? Second derivative test bending complete karta hai aur hidden inflection rule out karta hai.

Neeche ka figure — kya observe karna hai. Teal dashed line $y = x$ corner to corner run karta hai; do burnt-orange branches plum vertical wall $x = 0$ ke ek-ek taraf hain. Right branch plum minimum $(1, 2)$ tak dip karta hai slant line ke thoda upar (concave up), left branch teal maximum $(- 1, - 2)$ tak rise karta hai thoda neeche (concave down). Har branch origin se door jaate jaate $y = x$ ki taraf flatten hoti hai — yahi curve apne slant asymptote se "chipakna" hai.

Figure 3 — $f (x) = (x^{2} + 1) / x$ : slant asymptote $y = x$ , wall $x = 0$ , min $(1, 2)$ , max $(- 1, - 2)$ .

Verify: slant slope $m = lim_{x \to \infty} \frac{x ^{2} + 1}{x ^{2}} = 1$ , intercept $c = lim_{x \to \infty} \frac{1}{x} = 0$ ; $f (1) = 1 + 1 = 2$ , $f (- 1) = - 1 - 1 = - 2$ ; $f^{''} (1) = 2 > 0$ (min), $f^{''} (- 1) = - 2 < 0$ (max); $f^{'}$ ka $x = - 1$ par $+ \to -$ aur $x = 1$ par $- \to +$ change. (Machine-checked below.)

Example 5 — Ek curve jo apni horizontal asymptote cross karta hai · cell G

Step 1 — Horizontal asymptote. $lim_{x \to \pm \infty} \frac{x}{x ^{2} + 1} = lim \frac{1/ x}{1 + 1/ x ^{2}} = 0 \Rightarrow y = 0.$ Ye step kyun? Top ka degree $<$ bottom ka degree ⇒ curve infinity par $x$ -axis ke paas flatten ho jaata hai.

Step 2 — Crossing point. $f (x) = 0$ set karo: $\frac{x}{x ^{2} + 1} = 0 \Rightarrow x = 0$ . Curve origin se guzarta hai — jo $y = 0$ par hi lie karta hai. Toh yeh apni khud ki asymptote ko $x = 0$ par cross karta hai. Ye step kyun? Asymptote ek end-behaviour statement hai, barrier nahi; $f = 0$ solve karne se crossing locate hoti hai.

Step 3 — Turning points (bounded bump). $f^{'} (x) = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} = 0 \Rightarrow x = \pm 1.$ $f (1) = \frac{1}{2}$ (max), $f (- 1) = - \frac{1}{2}$ (min). Curve origin se rise karke $(1, \frac{1}{2})$ par ek gentle peak banata hai, phir $y = 0$ ki taraf wapas fall karta hai. Ye step kyun? Shape confirm karta hai: do humps ek crossing ke dono taraf origin par.

Neeche ka figure — kya observe karna hai. Teal dashed line $y = 0$ horizontal asymptote hai, aur burnt-orange curve origin par ink square se seedha cross karta hai — asymptote koi fence nahi hai. Curve plum maximum $(1, \frac{1}{2})$ tak rise karta hai, phir axis ki taraf wapas ease karta hai; oddness se yeh plum minimum $(- 1, - \frac{1}{2})$ tak left par mirror karta hai. Dono directions mein door jaate jaate yeh $y = 0$ par settle ho jaata hai, right se upar se aur left se neeche se.

Figure 4 — $f (x) = x / (x^{2} + 1)$ : crosses its horizontal asymptote $y = 0$ at the origin, max $(1, \frac{1}{2})$ , min $(- 1, - \frac{1}{2})$ .

Verify: $lim_{x \to \infty} f = 0$ ; $f (0) = 0$ (asymptote par); $f (1) = \frac{1}{1 + 1} = \frac{1}{2}$ , $f (- 1) = - \frac{1}{2}$ . (Machine-checked below.)

Example 6 — Degenerate input: removable hole vs blow-up · cell H

Step 1 — $g$ mein removable hole test karo. $x = 1$ plug karne se $\frac{0}{0}$ milta hai — yeh ek indeterminate form hai, automatically $\infty$ nahi. Factor karo: $g (x) = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1 (x \neq = 1) .$ $(x - 1)$ cancel ho jaata hai, toh $g$ sirf line $y = x + 1$ hai $x = 1$ par ek hole punch karke (woh value $2$ hoti). Ye step kyun? Limits ek $0/0$ ko alag karte hain jo simplify hota hai (removable) ek $\frac{nonzero}{0}$ se (true pole).

Step 2 — $h$ mein pole test karo (dono one-sided limits). $x = 1$ par, $h = \frac{1}{0}$ hai non-zero numerator ke saath, toh yeh cancel nahi ho sakta. Har side alag lelo:

Jaise $x \to 1^{+}$ , denominator $x - 1 \to 0^{+}$ (small positive), toh $h = \frac{1}{0 ^{+}} \to + \infty$ .
Jaise $x \to 1^{-}$ , denominator $x - 1 \to 0^{-}$ (small negative), toh $h = \frac{1}{0 ^{-}} \to - \infty$ .

Kyunki dono one-sided limits opposite infinities ki taraf jaate hain, $x = 1$ ek genuine two-sided vertical asymptote hai — curve wall ke right par upar shoot karta hai aur left par neeche. Ye step kyun? Nonzero over zero kabhi remove nahi ho sakta; har side par tiny denominator ka sign fix karta hai ki branch kis infinity ko chase karta hai.

Verify: $lim_{x \to 1} g (x) = 1 + 1 = 2$ (finite ⇒ removable hole; $g$ , $1$ par undefined hai lekin limit exist karti hai). $h$ ke liye: $lim_{x \to 1^{+}} h = + \infty$ , $lim_{x \to 1^{-}} h = - \infty$ (two-signed pole). (Machine-checked below.)

Example 7 — Word problem: cost-per-unit curve ki shape · cell I

Step 1 — Simplify (slant asymptote disguise mein spot karo). $A (x) = x + \frac{400}{x}, x > 0.$ Bade $x$ ke liye, $A \approx x$ (rising line); chhote $x$ ke liye, $\frac{400}{x}$ term dominate karta hai aur $A \to + \infty$ . Ek minimum beech mein hi hoga. Ye step kyun? Ye Example 4 ka structure ek real curve par apply hai; slant $y = x$ large-scale trend batata hai. Ye exactly Optimization mein study kiya gaya "find the lowest point" wala question hai.

Step 2 — First derivative se minimize karo. $A^{'} (x) = 1 - \frac{400}{x ^{2}} = 0 \Rightarrow x^{2} = 400 \Rightarrow x = 20 (positive root lo kyunki x > 0) .$ Check karo ki minimum hai: $A^{''} (x) = \frac{800}{x ^{3}} > 0$ for $x > 0$ → concave up → minimum. Ye step kyun? Optimization: ek smooth positive curve ka lowest point wahan hota hai jahan slope $= 0$ aur $A^{''} > 0$ .

Step 3 — Minimum value. $A (20) = 20 + \frac{400}{20} = 20 + 20 = 40 dollars .$ Ye step kyun? Physical question ka jawab deta hai: $20$ (hundred) items banana sabse sasta average cost deta hai, $40 each.

Verify: critical $x = 20$ ( $x^{2} = 400$ ka positive root); $A (20) = 20 + 20 = 40$ dollars/item; $A^{''} (20) = \frac{800}{2 0 ^{3}} = \frac{800}{8000} = \frac{1}{10} > 0$ (concave up ⇒ minimum). Units: dollars per item, poore mein consistent. (Machine-checked below.)

Example 8 — Exam twist: behaviour ek parameter par depend karta hai · cell J

Step 1 — Differentiate karo aur critical points locate karo. $f^{'} (x) = 3 x^{2} + a .$ $3 x^{2} + a = 0 \Rightarrow x^{2} = - \frac{a}{3}$ solve karo. Ye step kyun? Turning points sirf wahan exist karte hain jahan $f^{'} = 0$ ke real solutions hon. Kya $x^{2} = - \frac{a}{3}$ ke real roots hain yeh completely right-hand side ke sign par depend karta hai.

Step 2 — $a$ ke sign par case split (teeno regimes).

$a < 0$ : toh $- \frac{a}{3} > 0$ , toh $x = \pm - a /3$ do distinct real critical points hain — ek local max aur ek local min: genuine wiggle wali S-curve. Example $a = - 3$ se $x^{3} - 3 x$ wapas milta hai Example 1 se, roots $x = \pm 1$ par. Do turning points.
$a = 0$ : toh $f^{'} = 3 x^{2} \geq 0$ , sirf $x = 0$ par zero aur koi sign change nahi — yeh Example 2 ka flat (stationary) inflection $x^{3}$ hai. Zero turning points.
$a > 0$ : toh $- \frac{a}{3} < 0$ , toh $x^{2} = - \frac{a}{3}$ ka koi real solution nahi; waise bhi $f^{'} = 3 x^{2} + a \geq a > 0$ har jagah hai, toh curve strictly increasing hai koi flat spot ke bina. Zero turning points.

Ye step kyun? Sirf ek quantity $- \frac{a}{3}$ (positive, zero, ya negative) hinge hai; teeno bullets parameter ki har possible value exhaust karte hain.

Step 3 — Concluding summary.

Sign of $a$	$f^{'} = 0$ ke real roots	Turning points	Shape
$a < 0$	do ( $x = \pm - a /3$ )	2 (max & min)	wiggling S
$a = 0$	ek ( $x = 0$ ), koi sign change nahi	0	flat inflection cubic
$a > 0$	koi nahi	0	strictly increasing

Toh cubic $x^{3} + a x$ ka genuine "hill and valley" sirf tab hota hai jab $a < 0$ ; $a = 0$ par aur saare $a > 0$ ke liye woh monotonically climb karti hai. Transition value $a = 0$ hai. Ye step kyun? Reader ko exam mein quote karne ke liye ek-line rule deta hai: "do turning points iff $a < 0$ ."

Verify: $a = - 3$ se critical points $x = \pm 1$ milte hain (do real); $a = 0$ se sirf $x = 0$ milta hai with $f^{'} = 3 x^{2} \geq 0$ (koi sign change nahi, zero turning points); $a = 3$ se $f^{'} = 3 x^{2} + 3 \geq 3 > 0$ har jagah milta hai (koi real critical points nahi). (Machine-checked below.)

Recall Matrix ke across self-test

Har curve ko uski defining feature se match karo (answers neeche). $x^{3} - 3 x$ ::: do turning points, odd (cell A/F) $x^{3}$ ::: flat inflection, koi extremum nahi (cell B) $\frac{x ^{2}}{x ^{2} - 1}$ ::: horizontal + twin vertical asymptotes, even (C/D/F) $\frac{x ^{2} + 1}{x}$ ::: oblique asymptote $y = x$ (cell E) $\frac{x}{x ^{2} + 1}$ ::: origin par apni horizontal asymptote cross karta hai (cell G) $\frac{x ^{2} - 1}{x - 1}$ ::: $x = 1$ par removable hole, pole nahi (cell H)

Connections

Curve sketching — systematic approach — parent checklist jo ye examples exercise karte hain.
First derivative and monotonicity — turning points aur climb/fall Ex 1, 2, 4, 5, 8 mein.
Second derivative test — max vs min confirm karna Ex 1, 3, 4, 7 mein.
Rational functions and asymptotes — horizontal, vertical aur oblique walls Ex 3, 4, 5 mein.
Limits and continuity — hole vs blow-up Ex 6 mein.
Optimization — average-cost minimum Ex 7 mein.
Mean Value Theorem — theoretical reason ki $f^{'}$ ka sign trend dictate karta hai.