Applications — increasing - decreasing, local extrema (first derivative test): Visual walkthrough

4.1.28 · D2 · Maths › Calculus I — Limits & Derivatives › Applications — increasing - decreasing, local extrema (first

Hum yeh words use karenge. Aur kuch bhi shuru karne se pehle inhe clearly define kar lete hain.

Definition Woh words jinhe hume earn karna hai

$f (x)$ — horizontal position $x$ par curve ki height. Socho: jab tum $x$ steps daayein chale, tab road kitni upar hai.
$f^{'} (x)$ — position $x$ par slope: road ki steepness theek wahan par. Positive = daayein taraf upar jhukti hai, negative = daayein taraf neeche jhukti hai, zero = uss waqt flat hai.
Interval $I$ — $x$ -values ki ek stretch, jaise $- 2$ aur $5$ ke beech ke saare $x$ .
Arrow " $\to$ " patterns mein jaise " $+ \to -$ " ka matlab hai "jab hum critical point $c$ ke through left-to-right chalte hain, slope ka sign pehle wale se doosre mein badal jata hai." Yeh multiplication nahi hai; yeh ek before/after snapshot hai.
"Positive" aur "negative" ka matlab hai zero se zyada aur zero se kam; jab bhi hum kehte hain ki koi run ya quantity $> 0$ hai, hum strictly zero se zyada ka matlab lete hain.

Yahi poora vocabulary hai. Neechey sab kuch inhi se draw hota hai.

Step 1 — "Slope" actually kya measure karta hai?

KYA. Curve par do points choose karo, ek left wala position $a$ par aur ek right wala position $b$ par, jahan $a < b$ . Unke beech ka average slope rise divided by run hota hai: $average slope \frac{f ( b ) - f ( a )}{b - a} .$

$f (b) - f (a)$ — rise: $b$ par road $a$ se kitni zyada (ya kam) upar hai.
$b - a$ — run: hum kitna daayein chale. Kyunki $a < b$ , yeh hamesha $> 0$ hota hai (strictly zero se zyada).

KYUN. Slope woh tool hai "rising ya falling" ke liye, kyunki fraction ka sign uske upar aur neeche ke signs se decide hota hai. Neeche wala $b - a$ strictly positive locked hai, isliye sirf rise ka sign humein upar ya neeche batata hai. Yahi poore theorem ka seed hai.

PICTURE. Horizontal run aur vertical rise se bana right triangle — hypotenuse woh seedhi line hai jo do points ko jodhti hai.

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

Step 2 — Run ko shrink karo: average slope se $f^{'}$ tak

KYA. $b$ ko $a$ ki taraf slide karo. Jaise-jaise run $b - a$ tiny hota jaata hai, joining line door-door chord hona band ho jaati hai aur curve se chipak jaati hai — woh tangent line ban jaati hai, aur uska slope exactly $f^{'} (a)$ hota hai: $f^{'} (a) = lim_{b \to a} \frac{f ( b ) - f ( a )}{b - a} .$

$lim_{b \to a}$ — "dekhte hain ki fraction kya approach karta hai jab $b$ dheere-dheere $a$ tak creep karta hai".
$f^{'} (a)$ — result: road ki slope single point $a$ par.

KYUN. Hume steepness ek point par chahiye, kisi stretch pe average karke nahi. Limit hi woh ek tool hai jo two-point ratio ko one-point steepness mein turn karta hai — yeh answer karta hai "do points ke merge hone par average slope kya settle hota hai?"

PICTURE. Teen chords of shrinking run, tangent par flatten hote hue.

Step 3 — Mean Value Theorem: wapas jaane ka bridge

KYA. Mean Value Theorem kehta hai: kisi bhi stretch $[a, b]$ par jahan $f$ continuous hai (koi break nahi) aur andar differentiable hai (koi corner nahi), wahan kam se kam ek interior point $c$ hoga jahan tangent slope average slope ke barabar hoga: $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .$

$c$ — ek mystery point strictly $a$ aur $b$ ke beech; hum nahi jaante kahan hai, bas itna ki exist karta hai.
Left side $f^{'} (c)$ — ek instant slope (ek point).
Right side — Step 1 ka average slope (do points).

KYUN. Step 1 ne average slope diya; Step 2 ne instant slope diya. MVT woh bridge hai jo kehta hai ki yeh dono kahin na kahin equal hain. Yahi equality humein $f^{'}$ (instant) ke baare mein ek fact ko $f (b)$ vs $f (a)$ (heights) ke baare mein fact mein convert karne deta hai.

PICTURE. Chord aur curve ko $c$ par kiss karti ek parallel tangent.

Step 4 — Rearrange karo: heights slope ke sign ki baat maante hain

KYA. MVT ko run $b - a$ se multiply karo: $f (b) - f (a) = sign jise hum control karte hain f^{'} (c) \cdot > 0 (b - a) .$

KYUN. Ab rise $f (b) - f (a)$ ek product ke roop mein likha gaya hai. Ek factor $(b - a)$ hamesha $> 0$ hai. Isliye rise ka sign poori tarah $f^{'} (c)$ ke sign se decide hota hai. Yahi payoff hai: slope ke sign ko control karo, control karo ki road upar gayi ya neeche. Slope ka sirf isolated points par zero tak touch karna bhi kaafi hai, jab tak woh kabhi negative nahi hota.

Agar $f^{'} > 0$ hara jagah on $I$ : tab $f^{'} (c) > 0$ , product positive hai $\Rightarrow f (b) > f (a)$ → increasing.
Agar $f^{'} < 0$ hara jagah: product negative hai $\Rightarrow f (b) < f (a)$ → decreasing.

PICTURE. Ek sign multiplication table ledger ki tarah draw kiya gaya: (slope sign) $\times$ (positive run) = (rise sign).

Step 5 — Road turn around kahan kar sakti hai? Critical points

KYA. Peak ya valley wahan hoti hai jahan road rising band karke falling shuru karti hai (ya vice versa). Bilkul tip par, tangent ya to flat hoti hai ( $f^{'} (c) = 0$ ) ya broken hoti hai ( $f^{'} (c)$ exist nahi karta — ek corner). Aise kisi bhi tarah ka interior point (jiske dono taraf curve ho) ek critical point hota hai.

KYUN. Height ke rising se falling switch karne ke liye, slope ko zero se pass hona hoga — ya zero ke across jump karna hoga. Woh strictly positive rehte hue $+$ se $-$ nahi ja sakta. Isliye turnarounds critical points par trapped hain. Baaki sab jagah, Step 4 turn ko forbid karta hai.

PICTURE. Ek smooth hilltop (flat tangent) aur ek sharp corner (koi tangent nahi) — dono turnarounds hain.

Step 6 — First Derivative Test: turn ko padhna

KYA. Critical point $c$ par baito. Uske just left aur just right ka slope sign dekho (yaad karo $+ \to -$ ka matlab hai "sign $c$ se pehle $+$ tha, $c$ ke baad $-$ ho jaata hai"): $upar phir neeche + \to - \Rightarrow local max, neeche phir upar - \to + \Rightarrow local min .$

KYUN. Step 4 se, "left par $+$ " ka matlab hai road $c$ tak chadhi; "right par $-$ " ka matlab hai woh $c$ se door utri. Chadh-ke-phir-utarna exactly ek hilltop hai. Mirror pattern ek valley hai. Hume sirf har taraf $f^{'}$ ka sign chahiye tha — kuch aur nahi.

PICTURE. Ek hill labelled $+ \to -$ valley $- \to +$ ke paas, walking direction dikhate hue arrows.

Step 7 — Degenerate cases: same sign, aur true plateaus

KYA. Kya hoga agar $f^{'} = 0$ at $c$ lekin road dono taraf rise kare? Example: $f (x) = x^{3}$ at $x = 0$ , jahan $f^{'} (x) = 3 x^{2} \geq 0$ with only zero at $x = 0$ . $+ \to + (ya - \to -) \Rightarrow koi extremum nahi .$ Ek teesri possibility bhi hai: $f^{'} = 0$ poore sub-interval par (ek genuine flat plateau, e.g. ek constant stretch). Wahan slope us side par " $+$ " ya " $-$ " hai hi nahi — woh us stretch par exactly $0$ hai, isliye road level hai, aur phir bhi koi strict peak ya valley nahi hai.

KYUN. $x^{3}$ case mein road ne kabhi turn around nahi kiya — usne sirf pause kiya, jaise ek upward staircase par ek flat landing: pehle rising, baad mein rising, koi peak ya trough nahi. True plateau par road ek poori stretch par flat hai, isliye koi single point strictly highest ya lowest nahi hai. Dono mein, $f^{'}$ ka sign actually kabhi flip nahi hua, jo ki ek extremum signal karne ki ek hi cheez hoti. Yahi trap hai jise neeche Mistake A ke roop mein summarize kiya gaya hai: $f^{'} (c) = 0$ akela kaafi nahi — tumhe sign change dekhna hoga.

PICTURE. $x^{3}$ ka graph origin par flat hota hua lekin upar jaata hua — ek horizontal tangent bina kisi turn ke.

Step 8 — Worked landscape: $f (x) = x^{3} - 3 x$

Differentiate karne se pehle, humein ek earned tool chahiye.

KYA. Sab kuch ek real cubic par apply karo.

$f^{'} (x) = 3 x^{2} - 3$ . (Power rule on $x^{3}$ gives $3 x^{2}$ ; on $- 3 x = - 3 x^{1}$ gives $- 3$ ; constant terms would give $0$ .)
$3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ — do critical points.
Test $x = - 2$ : $f^{'} = 3 (4) - 3 = 9 > 0$ (increasing). Test $x = 0$ : $f^{'} = - 3 < 0$ (decreasing). Test $x = 2$ : $f^{'} = 9 > 0$ (increasing).
At $x = - 1$ : $+ \to -$ → local max, height $f (- 1) = (- 1)^{3} - 3 (- 1) = 2$ .
At $x = + 1$ : $- \to +$ → local min, height $f (1) = 1 - 3 = - 2$ .

KYUN. Yeh poora pipeline dikhata hai: differentiate karo → critical points dhundho → signs ko number-line karo → classify karo. Yeh Mistake D bhi drill karta hai: value report karo ( $2$ aur $- 2$ ), sirf location nahi.

PICTURE. Curve $x^{3} - 3 x$ neeche sign-of- $f^{'}$ number line ke saath, extrema marked.

Ek-picture summary

Upar sab kuch ek diagram mein collapse ho jaata hai: upar heights, neeche number line par slope sign map, turnarounds exactly wahan jahan sign flip hoti hai.

Recall Feynman retelling — poora walkthrough simple words mein

Socho tum ek hilly road par left se right chal rahe ho. Slope sirf yeh hai ki tumhare paanv ke neeche zameen kitni steep hai: uphill "positive" hai, downhill "negative". Agar mujhe ek exact jagah ki steepness chahiye, to main do nearby footprints leta hoon, rise-over-run measure karta hoon, phir unhe ek saath slide karta hoon jab tak woh same jagah na ho jaayein — woh limiting steepness $f^{'}$ hai. Mean Value Theorem ek promise hai: kisi bhi stretch par, kahin na kahin ground ki exact steepness poori stretch ki average steepness ke barabar hogi. Usse chali gayi (hamesha-positive) length se multiply karo, aur tum seekh lo ki tum upar gaye ya neeche — isliye slope ka sign akela decide karta hai rising ya falling; slope ka sirf scattered points par zero touch karna bhi kaafi hai, jab tak woh kabhi negative nahi hota. Turn around karne ke liye — hilltop ya valley — ground ko flat hona hoga ya kink hona hoga; woh critical points hain, aur ek fenced-off road ke far ends (endpoints) bhi candidates hain. Ek par khado aur dekhlo: chota arrow " $\to$ " ka matlab hai "pehle phir baad mein". Uphill-pehle, downhill-baad ( $+ \to -$ ) matlab tum peak par ho; downhill-pehle, uphill-baad ( $- \to +$ ) matlab ek valley hai. Lekin agar dono taraf uphill hai, ya poori stretch par dead flat hai, to tum sirf ek landing par se guzre — koi peak nahi. Isliye $f^{'} = 0$ akele (Mistake A) kabhi summit prove nahi karta; tumhe sign change dekhna hoga. Aur jab report karo, height $f (c)$ do, sirf spot $c$ nahi (Mistake D). Sab kuch $x^{3} - 3 x$ par test karo: $x = - 1$ par flat (peak, height $2$ ) aur $x = 1$ par (valley, height $- 2$ ), aur tumne poora landscape map kar liya.

Connections

Mean Value Theorem — Steps 3–4 iske direct consequence hain.
Fermat's Theorem on Stationary Points — kyun turnarounds interior critical points par trapped hain (Step 5).
Second Derivative Test — same critical points ko sign changes ki jagah concavity use karke classify karta hai.
Concavity and Inflection Points — Step 7 ka $x^{3}$ pause ek inflection hai.
Optimization (Closed Interval Method) — is local picture ko global extrema tak extend karta hai, including endpoints.
Curve Sketching — is walkthrough ko $f^{''}$ analysis ke saath combine karta hai.