Mean Value Theorem — proof, Rolle's theorem: Visual walkthrough

4.1.27 · D2 · Maths › Calculus I — Limits & Derivatives › Mean Value Theorem — proof, Rolle's theorem

Hum bilkul zero se derive karne wale hain: $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} for some c strictly inside (a, b) .$

Isse pehle ki hum woh line padh sakein, aao iske har piece ko build karein.

Step 1 — Secant kya hoti hai, aur woh fraction kya hai?

KYA. Figure mein do amber dots dekho: ek $(a, f (a))$ par, ek $(b, f (b))$ par. Unke through cyan seedhi line secant hai.

KYUN. "Average behaviour" ka matlab hamesha start-to-end hota hai, beech ke wiggles ignore karke. Secant exactly curve ka start-to-end straight-line summary hai — bilkul map par dashed "as-the-crow-flies" path ki tarah.

PICTURE. Jis fraction ki hum baat kar rahe hain woh us cyan line ki steepness hai: $secant slope \frac{f ( b ) - f ( a )}{b - a} = \frac{f ( b ) - f ( a ) rise: how much higher}{run: how far across b - a} .$

$f (b) - f (a)$ rise hai — figure ke right side par vertical amber segment.
$b - a$ run hai — bottom ke saath horizontal amber segment.
Unka ratio "rise over run" hai, slope ka standard matlab: ek unit across jane par kitne units upar.

Figure — Mean Value Theorem — proof, Rolle's theorem

Step 2 — Tangent kya hoti hai, aur $f^{'} (c)$ kya hai?

KYA. Figure mein cyan tangent curve ko ek point $c$ par kiss karti hai aur wahan uski tilt exactly match karti hai.

YEH TOOL KYUN — strictly ek limit kyun, slope kyun nahi? Secant ko rise/run compute karne ke liye do points chahiye. Tangent ek point par rehti hai, isliye ordinary "rise over run" $\frac{0}{0}$ hoga — undefined. Derivative exactly woh tool hai jo "ek point se slope kya hai?" ka jawab dene ke liye bana hai. Yeh dusre point $c + h$ ko $c$ ki taraf slide karke aur secant slope settle hote dekhkar karta hai: $f^{'} (c) = lim_{h \to 0} \frac{f ( c + h ) - f ( c )}{h} .$

$h$ $c$ se ek tiny step sideways hai (figure mein amber bracket).
$f (c + h) - f (c)$ us step par tiny rise hai.
$lim_{h \to 0}$ matlab "step ko kuch nahi tak shrink hone do aur woh number report karo jis par ratio approach karta hai."

PICTURE. Figure mein pale secants ko dekho jo $h \to 0$ hone par bold cyan tangent ki taraf fan hoti hain.

Step 3 — Trick: ends ko flatten karne ke liye secant subtract karo

KYA. Har $x$ ke liye, curve se secant tak ek vertical amber line girano. Uski signed length $g (x)$ hai: positive jahan curve line ke upar hai, negative jahan neeche. Figure mein kuch aisi gap sticks dikhayi gayi hain.

YEH STEP KYUN. Rolle's theorem (aasaan, flat-secant case) ko dono endpoints same height par chahiye. Hamare curve ke endpoints generally same height par nahi hote. Isliye hum tilt subtract kar dete hain: secant hi tilt hai, aur use hataane par dono ends height $0$ par aa jaate hain. Hum wiggles ki shape nahi badal rahe — ek seedhi line subtract karna har point ko vertically shift karta hai lekin peaks ko peaks aur valleys ko valleys rakhta hai.

$L (x)$ ke andar term by term:

$f (a)$ — starting height (taaki line $x = a$ par exactly curve par shuru ho).
$\frac{f ( b ) - f ( a )}{b - a}$ — Step 1 ka secant slope.
$(x - a)$ — left post se hum kitna right chale; slope se multiply karne par woh extra height milti hai jo line ne climb ki hai.

Step 4 — Check karo ki gap ke endpoints equal hain (dono zero hain)

KYA. $g$ ko dono posts par compute karo.

KYUN. Yeh teen Rolle hypotheses hain jinhe hum $g$ ke liye confirm karna chahte hain $g$ par Rolle use karne ki permission se pehle. Yahan hum hypothesis 3 (equal endpoint heights) nail karte hain.

$g (a) = f (a) - L (a) [f (a) + \frac{f ( b ) - f ( a )}{b - a} (a - a)] = f (a) - f (a) = 0.$

$(a - a) = 0$ slope term ko kill kar deta hai, isliye $L (a) = f (a)$ : line curve par shuru hoti hai.

$g (b) = f (b) - L (b) [f (a) + \frac{f ( b ) - f ( a )}{b - a} (b - a)] = f (b) - [f (a) + (f (b) - f (a))] = f (b) - f (b) = 0.$

Numerator aur denominator mein $(b - a)$ cancel ho jaata hai, $f (a) + f (b) - f (a) = f (b)$ bachta hai: line curve par bhi khatam hoti hai.

PICTURE. Figure mein ends par dono amber gap-sticks collapse hokar length zero ho gayi hain — curve $g$ $a$ aur $b$ dono par axis par pin ho jaati hai.

Step 5 — Rolle ek flat spot guarantee karta hai: woh kahaan se aata hai?

KYA (engine). Peak ya valley zero slope kyun force karti hai? Do named tools kaam karte hain:

Extreme Value Theorem — ek closed interval $[a, b]$ par continuous function kahin maximum height $M$ aur minimum height $m$ attain karta hai. Isliye peak/valley genuinely exist karti hai (yeh koi mirage nahi jise hum approach karte hain lekin reach nahi karte).
Fermat's Theorem (interior extrema) — agar woh highest/lowest point strictly inside $(a, b)$ baitha ho, to wahan derivative $0$ hai.

Extremum interior kyun hai (woh case jo break ho sakta tha, usse cover karna). Agar dono $M$ aur $m$ sirf endpoints par hote, to $M = g (a) = 0 = g (b) = m$ , isliye $M = m$ — matlab $g$ constant zero hai (flat case, upar handle kiya). Warna $M, m$ mein se kam se kam ek aisi value hai jo endpoints share nahi karte, isliye use ek interior point $c \in (a, b)$ par achieve kiya jana chahiye.

Fermat ise zero par kyun pin karta hai — squeeze. Sit at an interior peak $c$ . Tiny $h > 0$ se right step karo: curve upar nahi ho sakti, isliye $g (c + h) - g (c) \leq 0$ , aur $h > 0$ se divide karne par, $\frac{g ( c + h ) - g ( c )}{h} \leq 0 \Rightarrow g^{'} (c) = lim_{h \to 0^{+}} (\dots) \leq 0.$ Tiny $h < 0$ se left step karo: phir $g (c + h) - g (c) \leq 0$ , lekin ab negative $h$ se divide karne par inequality flip ho jaati hai, $\frac{g ( c + h ) - g ( c )}{h} \geq 0 \Rightarrow g^{'} (c) = lim_{h \to 0^{-}} (\dots) \geq 0.$ Kyunki $g$ differentiable hai, dono one-sided limits same number $g^{'} (c)$ ke barabar hain. Jo number $\leq 0$ aur $\geq 0$ ho woh sirf $0$ ho sakta hai: $g^{'} (c) = 0.$

PICTURE. Figure $g$ ke interior peak ko amber dot se mark karti hai; wahan tangent ek horizontal cyan line hai — flat spot.

Step 6 — Untilt: " $g$ flat" ko MVT equation mein badlo

KYA. Definition $g (x) = f (x) - L (x)$ differentiate karo aur flat point daalo.

KYUN. $g$ ka flat spot $f$ ke baare mein ek hidden statement hai. Ise reveal karne ke liye subtraction undo karo.

$g^{'} (x) = f^{'} (x) - slope of L, a constant \frac{f ( b ) - f ( a )}{b - a} .$

$f^{'} (x)$ — original curve ka slope.
Secant slope ek fixed number hai, isliye uska derivative wahi constant har jagah contribute karta hai (ek seedhi line ka slope kabhi nahi badlata).

Ab Step 5 ka special point $c$ plug in karo, jahan $g^{'} (c) = 0$ hai: $0 = f^{'} (c) - \frac{f ( b ) - f ( a )}{b - a} ⟹ f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .$

PICTURE. Figure mein $g$ ka horizontal tangent (Step 5 se) secant slope se wapas upar tilt hota hai. Secant ki steepness se tilt ki gayi flat line exactly secant ki steepness wali line ban jaati hai — $f$ ki cyan tangent ab amber secant ke parallel run karti hai. Woh parallelism hi MVT hai.

Step 7 — Degenerate aur edge cases (reader ko koi na-dikhaya scenario mat milne do)

Worked example Case-by-case, taaki kuch bhi uncovered na rahe

Case A — $f (a) = f (b)$ (Rolle khud). Tab secant slope $\frac{0}{b - a} = 0$ hai. MVT conclusion ban jaata hai $f^{'} (c) = 0$ : ek horizontal tangent. Rolle literally MVT hai level secant ke saath.

Case B — $f$ constant. Har point flat spot hai; $f^{'} (x) = 0$ sabhi $x$ ke liye, aur secant slope bhi $0$ hai. Equation $0 = 0$ har interior $c$ par hold hoti hai — infinitely many valid choices.

Case C — $f$ ek seedhi line. Tab $f^{'}$ everywhere ek fixed number hai, aur woh already secant slope ke barabar hai. Har interior point $c$ ki tarah kaam karta hai.

Case D — kai flat spots (ek wavy curve). MVT kehta hai "kam se kam ek" $c$ , "exactly ek" kabhi nahi. Ek curve jo secant slope ko repeatedly cross karti hai uske paas kai valid $c$ 's hain. Dekho Increasing and Decreasing Functions link — $f^{'}$ ke sign changes multiple crossings banate hain.

Case E — hypotheses violate hain. Agar $f$ mein corner ho (differentiable nahi) jaise $∣ x ∣$ $[- 1, 1]$ par, to interior flat tangent exist karne ki zaroorat nahi; guarantee khatam ho jaati hai. Continuity/differentiability load-bearing hain, decorative nahi.

Ek-picture summary

Sab kuch ek frame mein: original curve $f$ apni amber secant ke saath; tilted gap curve $g$ dono ends par $0$ par pinned; $g$ ka flat spot (open circle) directly upar $f$ ki parallel tangent (cyan) ke. Unhe connect karne wali vertical guide line follow karo — woh single vertical line hi proof hai: flat $g$ upar ⇔ secant-parallel $f$ neeche.

Recall Feynman retelling — plain words mein poora walkthrough

Tum prove karna chahte ho: kahin tumhari steepness tumhari average steepness se match ki. Dikkat yeh hai ki "average" do door-door points compare karta hai, jabki "abhi steepness" ek single point par rehti hai — line up karna awkward hai.

Isliye ek trick khelo. Poora landscape itna tip karo ki tumhara start aur finish same height par baith jaayein (yahi secant subtract karna hai — amber seedhi line). Tip karna hills ko erase nahi karta; sirf unke neeche zameen re-level ho jaati hai. Ab tum same height par shuru aur khatam karte ho, isliye jaise ek walk jo tumhare dost ke level par wapas aati hai, ek moment hota hai jab tum perfectly flat move kar rahe ho — kisi hill ki top ya dip ki bottom par (yahi Rolle hai, "top exist karta hai" se Extreme Value Theorem aur "top flat hai" se Fermat se powered).

Aakhir mein, landscape ko wapas waisa hi tip karo jaisa actually tha. Woh flat moment exactly utna tilt ho jaata hai jitna tumne tip kiya tha — average steepness. Isliye us instant par tumhari real steepness average ke barabar hai. Ho gaya.

Recall One-line memory hooks

Secant slope :::: $\frac{f ( b ) - f ( a )}{b - a}$ , average steepness (rise over run). Gap function $g$ :::: $f (x) - L (x)$ , curve se secant tak vertical distance. $g (a) = g (b) = 0$ kyun :::: secant subtract karna dono endpoints ko line par bitha deta hai. Engine of the flat spot :::: Extreme Value Theorem (top exist karta hai) + Fermat (top flat hai). Untilt step :::: $g^{'} (c) = 0 \Rightarrow f^{'} (c) =$ secant slope.

Connections

Mean Value Theorem — proof, Rolle's theorem — woh parent jise yeh page visualize karta hai.
Extreme Value Theorem — guarantee karta hai ki $g$ ka peak/valley actually exist karta hai (Step 5).
Fermat's Theorem (interior extrema) — interior peak ⇒ $f^{'} = 0$ , Step 5 mein squeeze.
Continuity · Differentiability — $g$ ke liye inherited hypotheses (Step 4).
Increasing and Decreasing Functions — multiple $c$ 's explain karta hai (Case D).
Cauchy's Mean Value Theorem — same "subtract, apply Rolle" trick do functions ke saath.
Taylor's Theorem — MVT uska degree-one seed hai.
Lipschitz Continuity — MVT ek derivative bound ko slope bound mein badalta hai (parent Example 3).