Applications — increasing - decreasing, local extrema (first derivative test): Worked examples

4.1.28 · D3 · Maths › Calculus I — Limits & Derivatives › Applications — increasing - decreasing, local extrema (first

Scenario matrix

Is topic ke har problem ka type inn cells mein se kisi ek mein aata hai. Aane wale examples mein woh cell label ki gayi hai jo woh cover karta hai, taaki end tak koi scenario na chhoote.

Cell	Situation	Tricky kya hai	Example
C1	Smooth polynomial, sign flip karta hai $+ \to -$ aur $- \to +$	standard peak aur valley	Ex 1
C2	Stationary point mein koi sign change nahi	$f^{'} (c) = 0$ par dono taraf same sign	Ex 2
C3	Derivative undefined (corner)	test ek kink par bhi kaam karta hai	Ex 3
C4	Derivative undefined (cusp, vertical tangent)	$f^{'} \to \pm \infty$ , phir bhi ek critical point	Ex 4
C5	Har jagah Monotone (koi critical points nahi)	confidently "no extrema" report karna hoga	Ex 5
C6	Rational function — $f^{'}$ undefined hai ek aisi jagah jo domain mein nahi	woh point critical NAHI hai	Ex 6
C7	Trig / periodic — infinitely many critical points	ek poori family classify karo	Ex 7
C8	Word problem (real-world max)	words → $f$ mein translate karo, phir test karo	Ex 8
C9	Exam twist — parameter $a$ answer decide karta hai	ek constant par case-split karo	Ex 9

Worked example Example 1 — cell C1: smooth quartic, ek valley–peak–valley

$f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2}$ ke har local extremum ko classify karo.

Forecast: Ek quartic upar ki taraf open hoti hai — guess karo: kitne turning points hain, aur woh max hain ya min?

Step 1 — differentiate karo. $f^{'} (x) = 12 x^{3} - 24 x^{2} + 12 x$ . Yeh step kyun? $f^{'}$ ki sign hi humara rising/falling ka poora map hai.

Step 2 — critical points dhundhne ke liye factor karo. $f^{'} (x) = 12 x (x^{2} - 2 x + 1) = 12 x (x - 1)^{2} .$ Toh $f^{'} (x) = 0$ at $x = 0$ aur $x = 1$ par. Dono domain mein hain (saare reals), isliye dono critical hain. Yeh step kyun? Factoring roots aur unki multiplicity expose karta hai — squared factor $(x - 1)^{2}$ ek warning hai ki $f^{'}$ zero ko touch kar sakta hai bina cross kiye.

Step 3 — har factor ki sign. $12 x$ , $x < 0$ ke liye negative hai, $x > 0$ ke liye positive. Factor $(x - 1)^{2} \geq 0$ hamesha (ek square kabhi negative nahi hota). Toh $f^{'}$ ki sign sirf $x$ ki sign hai, bas yeh $x = 1$ par $0$ tak gir jaata hai bina sign change kiye. Yeh step kyun? Factor-by-factor padhna, squares hone par blind test points se better hai.

Step 4 — number line (figure dekho).

Interval	test $x$	$12 x$	$(x - 1)^{2}$	$f^{'}$	behaviour
$(- \infty, 0)$	$- 1$	$-$	$+$	$-$	decreasing
$(0, 1)$	$0.5$	$+$	$+$	$+$	increasing
$(1, \infty)$	$2$	$+$	$+$	$+$	increasing

Step 5 — classify karo.

$x = 0$ par: $- \to +$ → local minimum, value $f (0) = 0$ .
$x = 1$ par: $+ \to +$ → koi extremum nahi (ek horizontal-tangent inflection, bilkul C2 ki warning).

Verify: $f (0) = 0$ . Original $f$ mein $0$ ke dono taraf ek point test karo: $f (- 1) = 3 + 8 + 6 = 17 > 0$ aur $f (0.5) = 3 (0.0625) - 8 (0.125) + 6 (0.25) = 0.1875 - 1 + 1.5 = 0.6875 > 0$ . Dono $f (0) = 0$ se upar hain ✓ — yeh sach mein ek valley ka bottom hai.

Worked example Example 2 — cell C2: stationary hai par extremum NAHI (isolated)

$f (x) = x^{5}$ ka, agar koi ho, local extremum kahan hai?

Forecast: $f^{'} (0) = 0$ , toh ek beginner "extremum!" chilla deta hai. Kya sach mein hai?

Step 1 — differentiate karo. $f^{'} (x) = 5 x^{4}$ . Yeh step kyun? Hume sirf ek candidate ke aas-paas sign map chahiye.

Step 2 — critical point. $5 x^{4} = 0 \Rightarrow x = 0$ (sirf ek root). Yeh step kyun? $x^{4}$ ek fourth power hai — yeh har jagah $\geq 0$ hai aur sirf $x = 0$ par $0$ hota hai.

Step 3 — sign check. $5 x^{4} > 0$ har $x \neq = 0$ ke liye. Toh $f^{'}$ , $0$ ke across $+ \to +$ hai. Yeh step kyun? Dono taraf same sign parent ka "neither" case hai.

Step 4 — conclude karo. $x = 0$ ek critical point hai par extremum nahi (horizontal tangent inflection, $x^{3}$ ki tarah).

Verify: $f (- 0.1) = - 1 0^{- 5} < 0 < 1 0^{- 5} = f (0.1)$ jabki $f (0) = 0$ beech mein baith jaata hai — function $0$ se guzarte hue chadhta rehta hai, koi peak ya valley nahi ✓.

Worked example Example 3 — cell C3: ek corner par derivative undefined

$f (x) = ∣ x - 2∣ + 1$ ke extrema classify karo.

Forecast: Ek "V" right aur upar shift ki hui. Uski tip kahan hai, aur kya woh max hai ya min?

Step 1 — piecewise derivative. $f (x) = {(x - 2) + 1 = x - 1, - (x - 2) + 1 = 3 - x, x \geq 2 x < 2$ toh $f^{'} (x) = + 1$ for $x > 2$ aur $f^{'} (x) = - 1$ for $x < 2$ ; $x = 2$ par left slope $- 1$ aur right slope $+ 1$ agree nahi karte, isliye $f^{'} (2)$ exist nahi karta. Yeh step kyun? Ek corner ka koi ek single tangent nahi hota — yahi precisely ek "undefined" type ka critical point hai (parent ki Mistake B).

Step 2 — sirf ek critical point hai $x = 2$ .

Step 3 — sign change. $x = 2$ ke across $- \to +$ . Yeh step kyun? Pehle neeche phir upar = valley.

Step 4 — conclude karo. $x = 2$ par local (aur global) minimum, value $f (2) = 0 + 1 = 1$ .

Verify: $f (1) = ∣ - 1∣ + 1 = 2$ aur $f (3) = ∣1∣ + 1 = 2$ ; dono $f (2) = 1$ se zyada hain ✓ — tip sabse nichla point hai.

Worked example Example 4 — cell C4: vertical tangent wala cusp

$f (x) = x^{2/3}$ ke extrema classify karo.

Forecast: Square-ish power ka even root — isko sketch karo, sharp point kahan hai?

Step 1 — differentiate karo. $f^{'} (x) = \frac{2}{3} x^{- 1/3} = \frac{2}{3 3 x}$ . Yeh step kyun? Power rule; negative exponent clearly dikhne do taaki hum dekh sakein yeh kahan blow up karta hai.

Step 2 — $f^{'}$ kahan zero ya undefined hai? Fraction $\frac{2}{3 3 x}$ kabhi zero nahi hota (numerator $2$ hai). Yeh $x = 0$ par undefined hai ( $30 = 0$ se division). Kyunki $x = 0$ $f$ ke domain mein hai, yeh ek critical point hai. Yeh step kyun? "Undefined derivative par point domain mein" = critical (Ex 6 se contrast karo, jahan bad point domain mein NAHI hai).

Step 3 — $f^{'}$ ki sign. $3 x$ , $x < 0$ ke liye negative hai, $x > 0$ ke liye positive. Toh

$x < 0$ : $f^{'} = \frac{+}{neg} < 0$ → decreasing.
$x > 0$ : $f^{'} = \frac{+}{pos} > 0$ → increasing.

Step 4 — conclude karo. $x = 0$ par $- \to +$ → local minimum, value $f (0) = 0$ . Note karo ki $f^{'} \to - \infty$ as $x \to 0^{-}$ aur $+ \infty$ as $x \to 0^{+}$ : dono branches ek cusp mein shoot up hoti hain, phir bhi sign test ise perfectly classify karta hai.

Verify: $f (- 8) = (- 8)^{2/3} = (3 - 8)^{2} = (- 2)^{2} = 4$ aur $f (8) = 4$ ; dono $f (0) = 0$ se upar hain ✓.

Worked example Example 5 — cell C5: har jagah monotone, koi extrema nahi

Dikhao ki $f (x) = x^{3} + x$ ka koi local extrema nahi hai, aur uski behaviour bolo.

Forecast: Ek cubic — kya uske turning points hone hi chahiye?

Step 1 — differentiate karo. $f^{'} (x) = 3 x^{2} + 1$ . Yeh step kyun? Hum sign changes dhundhte hain.

Step 2 — $f^{'} = 0$ solve karo. $3 x^{2} + 1 = 0 \Rightarrow x^{2} = - \frac{1}{3}$ , jiska koi real solution nahi (ek square negative nahi ho sakta). Aur $f^{'}$ har jagah defined hai. Toh koi bhi critical points nahi hain. Yeh step kyun? Koi critical point nahi ⇒ sign flip hone ki koi jagah nahi.

Step 3 — $f^{'}$ ki sign. $3 x^{2} \geq 0$ , toh $3 x^{2} + 1 \geq 1 > 0$ har $x$ ke liye. Isliye $f^{'} > 0$ har jagah. Yeh step kyun? Ek unbroken sign confirm karta hai.

Step 4 — conclude karo. $f$ $R$ par strictly increasing hai; koi local maximum ya minimum exist nahi karta. Yeh parent ke Mean Value Theorem argument ka saaf use hai: $f^{'} > 0$ poori line par ⇒ throughout increasing.

Verify: koi bhi $a < b$ lo, jaise $0 < 1$ : $f (0) = 0 < 2 = f (1)$ ✓; aur $- 1 < 0$ : $f (- 1) = - 2 < 0 = f (0)$ ✓ — kabhi turn nahi karta.

Worked example Example 6 — cell C6: rational function, "bad" point domain mein NAHI

$f (x) = \frac{x ^{2}}{x - 1}$ ke extrema classify karo.

Forecast: $x = 1$ par trouble hai. Kya $x = 1$ ek critical point hai? (Dhyan raho!)

Step 1 — pehle domain. $x = 1$ denominator ko $0$ bana deta hai, isliye $x = 1$ domain mein nahi hai. Yeh kabhi ek critical point ya extremum nahi ho sakta — extrema domain points hone chahiye. Yeh step kyun? Isko skip karna classic trap hai; domain se bahar ek point ek vertical asymptote hai, peak nahi.

Step 2 — differentiate karo (quotient rule). $f^{'} (x) = \frac{( 2 x ) ( x - 1 ) - x ^{2} ( 1 )}{( x - 1 ) ^{2}} = \frac{2 x ^{2} - 2 x - x ^{2}}{( x - 1 ) ^{2}} = \frac{x ^{2} - 2 x}{( x - 1 ) ^{2}} = \frac{x ( x - 2 )}{( x - 1 ) ^{2}} .$ Yeh step kyun? Quotient rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

Step 3 — critical points. Numerator zero: $x (x - 2) = 0 \Rightarrow x = 0$ ya $x = 2$ (dono domain mein). Denominator $(x - 1)^{2} > 0$ except $x = 1$ par, jo already exclude hai. Toh critical points $x = 0, x = 2$ hain.

Step 4 — $f^{'}$ ki sign (denominator hamesha $+$ hai, isliye sign $=$ $x (x - 2)$ ki sign):

Interval	$x (x - 2)$	$f^{'}$	behaviour
$(- \infty, 0)$	$(-) (-) = +$	$+$	increasing
$(0, 1)$	$(+) (-) = -$	$-$	decreasing
$(1, 2)$	$(+) (-) = -$	$-$	decreasing
$(2, \infty)$	$(+) (+) = +$	$+$	increasing

( $x = 1$ ke across sign flip nahi karta; hum us excluded point ko skip kar dete hain.)

Step 5 — classify karo.

$x = 0$ : $+ \to -$ → local maximum, value $f (0) = \frac{0}{- 1} = 0$ .
$x = 2$ : $- \to +$ → local minimum, value $f (2) = \frac{4}{1} = 4$ .

Verify: $f (- 1) = \frac{1}{- 2} = - 0.5 < 0 = f (0)$ aur $f (0.5) = \frac{0.25}{- 0.5} = - 0.5 < 0$ , toh $0$ apne neighbours se upar hai ✓. $f (1.5) = \frac{2.25}{0.5} = 4.5 > 4 = f (2)$ aur $f (3) = \frac{9}{2} = 4.5 > 4$ , toh $4$ valley ka floor hai ✓.

Worked example Example 7 — cell C7: periodic function, infinitely many critical points

$f (x) = sin x + cos x$ ke local extrema $R$ par dhundho aur classify karo.

Forecast: Waves repeat karte hain — ek pattern of alternating peaks aur troughs expect karo, sirf ek nahi.

Step 1 — differentiate karo. $f^{'} (x) = cos x - sin x$ . Yeh step kyun? $f^{'}$ ki sign rule karti hai, chahe functions periodic hon.

Step 2 — $f^{'} = 0$ solve karo. $cos x = sin x \Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4} + nπ$ har integer $n$ ke liye. Yeh step kyun? $tan x = 1$ har $π$ par repeat karta hai, infinite family deta hai. $f^{'}$ har jagah defined hai, isliye yahi sirf critical points hain.

Step 3 — ek period par sign test. $n = 0$ , $x = \frac{π}{4}$ lo. Bilkul left ( $x = 0$ ): $f^{'} = cos 0 - sin 0 = 1 > 0$ . Bilkul right ( $x = \frac{π}{2}$ ): $f^{'} = 0 - 1 = - 1 < 0$ . Toh $+ \to -$ → $x = \frac{π}{4}$ par local maximum, value $f (\frac{π}{4}) = \frac{2}{2} + \frac{2}{2} = 2$ . $n = 1$ , $x = \frac{5 π}{4}$ lo: left ( $x = π$ ) $f^{'} = - 1 - 0 = - 1 < 0$ , right ( $x = \frac{3 π}{2}$ ) $f^{'} = 0 + 1 = 1 > 0$ , toh $- \to +$ → local minimum, value $f (\frac{5 π}{4}) = - \frac{2}{2} - \frac{2}{2} = - 2$ . Yeh step kyun? Ek period poore pattern ko periodicity se settle kar deta hai.

Step 4 — rule. $x = \frac{π}{4} + 2 nπ$ par Maxima (value $2$ ), $x = \frac{5 π}{4} + 2 nπ$ par minima (value $- 2$ ). Yeh hamesha ke liye alternate karte hain.

Verify: $sin x + cos x = 2 sin (x + \frac{π}{4})$ ka amplitude $2$ hai, toh true max $2$ aur min $- 2$ hai ✓ — match karta hai.

Worked example Example 8 — cell C8: word problem (real-world maximum)

Ek farmer ke paas ek seedhi nadi ke sath ek rectangular pen banane ke liye $200$ m fence hai (nadi ki taraf koi fence nahi chahiye). Kaunse dimensions sabse bada area denge?

Forecast: Fixed perimeter ke saath, guess karo: kya ek long-thin pen zyada badi hogi ya ek squarish pen?

Step 1 — function mein translate karo. $x$ = nadi ke perpendicular dono sides mein se har ek ki length, aur $y$ = nadi ke parallel single side. Fence use: $2 x + y = 200 \Rightarrow y = 200 - 2 x$ . Area $A = x y = x (200 - 2 x) = 200 x - 2 x^{2}$ , valid for $0 \leq x \leq 100$ . Yeh step kyun? Derivative test run hone se pehle hume ek single-variable function lena hoga.

Step 2 — differentiate karo. $A^{'} (x) = 200 - 4 x$ . Yeh step kyun? $A^{'}$ ki sign batati hai kahan area rise/fall karta hai.

Step 3 — critical point. $200 - 4 x = 0 \Rightarrow x = 50$ . Har jagah defined, ek critical point.

Step 4 — sign test. $A^{'} (40) = 200 - 160 = 40 > 0$ (increasing), $A^{'} (60) = 200 - 240 = - 40 < 0$ (decreasing). $+ \to -$ → $x = 50$ par local maximum. Yeh step kyun? First derivative test prove karta hai ki yeh sach mein ek peak hai, min ya inflection nahi.

Step 5 — dimensions & area. $x = 50$ m, $y = 200 - 2 (50) = 100$ m, area $A = 50 \cdot 100 = 5000 m^{2}$ . Dono report karo: location $x = 50$ aur value $5000 m^{2}$ (parent ki Mistake D).

Verify: Endpoints dete hain $A (0) = 0$ aur $A (100) = 100 (0) = 0$ ; interior point $A (50) = 5000$ dono se better hai, aur units m·m $= m^{2}$ hain ✓.

Worked example Example 9 — cell C9: exam twist, answer ek parameter par depend karta hai

Constant $a$ ki kin values ke liye $f (x) = x^{3} + a x$ ke local extrema hain, aur kahan?

Forecast: Guess karo: kya har choice of $a$ turning points deti hai, ya sirf kuch?

Step 1 — differentiate karo. $f^{'} (x) = 3 x^{2} + a$ . Yeh step kyun? Critical points $a$ par depend karte hain, isliye hum ise symbolic rakhte hain.

Step 2 — $f^{'} = 0$ solve karo. $3 x^{2} + a = 0 \Rightarrow x^{2} = - \frac{a}{3}$ . Real solution tabhi exist karta hai jab $- \frac{a}{3} \geq 0$ , yaani $a \leq 0$ . Yeh step kyun? $x^{2} \geq 0$ , isliye right side non-negative hona chahiye — yeh problem ko cases mein split karta hai.

Step 3 — case split.

$a > 0$ : koi real critical point nahi ⇒ $f^{'} = 3 x^{2} + a > 0$ har jagah ⇒ strictly increasing, koi extrema nahi (Ex 5 ki tarah).
$a = 0$ : $f = x^{3}$ , single critical point $x = 0$ jahan $f^{'} = 3 x^{2} \geq 0$ , same sign ⇒ koi extremum nahi (Ex 2 ki tarah).
$a < 0$ : do critical points $x = \pm - a /3$ . Kyunki $3 x^{2} + a$ , $x$ mein ek upward parabola hai, yeh roots ke bahar $+$ aur unke beech $-$ hai. Toh left root par $+ \to -$ = local max, right root par $- \to +$ = local min. Yeh step kyun? Matrix ki har cell ek case ke roop mein wapas aati hai jo $a$ se control hota hai.

Step 4 — concrete check. $a = - 3$ ke liye: roots $x = \pm 1 = \pm 1$ ; yeh exactly parent ka Example 1 hai ( $x^{3} - 3 x$ ): $- 1$ par max, $+ 1$ par min.

Verify: $a = - 3$ ke saath, $f (- 1) = - 1 + 3 = 2$ (max value) aur $f (1) = 1 - 3 = - 2$ (min value) ✓ — parent note se match karta hai. $a = 3$ ke saath: $f^{'} = 3 x^{2} + 3 \geq 3 > 0$ , monotone, koi turning point nahi ✓.

Recall Active recall — answers cover karo

Ex 1 mein, $x = 1$ extremum kyun nahi hai? ::: $f^{'} = 12 x (x - 1)^{2}$ ; $(x - 1)^{2}$ factor $f^{'}$ ko dono taraf positive rakhta hai — same sign.
Ex 6 mein, $x = 1$ critical point kyun NAHI hai? ::: $x = 1$ domain mein nahi hai (denominator zero) — extrema domain points hone chahiye.
Ex 4: $x^{2/3}$ ke liye $x = 0$ kis tarah ka point hai? ::: Ek cusp / local minimum; $f^{'}$ wahan undefined hai par $- \to +$ .
Ex 9: kin $a$ ke liye $x^{3} + a x$ ke extrema hain? ::: Sirf $a < 0$ ke liye; tab $- - a /3$ par max, $+ - a /3$ par min.
Ex 8: 200 m fence nadi ke sath max area? ::: $x = 50$ , $y = 100$ , area $5000 m^{2}$ .
Ex 7: $sin x + cos x$ ke extrema? ::: Max $2$ at $\frac{π}{4} + 2 nπ$ , min $- 2$ at $\frac{5 π}{4} + 2 nπ$ .

Connections

Fermat's Theorem on Stationary Points — kyun hum sirf critical points dhundhe hain (Ex 6 ka domain caveat).
Mean Value Theorem — Ex 5 mein monotone conclusion ke peeche ka engine.
Second Derivative Test — jab apply hota hai to faster classifier hai; Ex 4 mein cusps aur Ex 2 mein $x^{5}$ par fail karta hai.
Optimization (Closed Interval Method) — Ex 8 global answer ke liye endpoints ke saath.
Curve Sketching — in saare sign charts ko full graphs mein assemble karna.
Concavity and Inflection Points — Ex 1 ke $x = 1$ aur Ex 2 ke $x = 0$ ki inflection nature.

Worked examples — Applications — increasing - decreasing, local extrema (first derivative test)

Scenario matrix

Connections

Concept Map