Mean Value Theorem — proof, Rolle's theorem: Worked examples

4.1.27 · D3 · Maths › Calculus I — Limits & Derivatives › Mean Value Theorem — proof, Rolle's theorem

Yeh page ek drill ground hai. Parent note ne prove kiya tha ki Mean Value Theorem (MVT) aur Rolle's theorem kyun sach hain. Yahan hum har tarah ki situation dhundhte hain jo exam mein aa sakti hai — simple curves, hidden traps wali curves, degenerate cases, word problems, aur exam twists — aur har ek ko zero se solve karte hain.

Do symbols jo poore time use honge, ek baar define kar lete hain taaki baad mein surprise na ho:

Scenario matrix

Is topic ka har problem in cells mein se kisi ek mein aata hai. Neeche ke examples mein cell label diya gaya hai.

Cell	Kya special hai	Example
A. Plain MVT	saari hypotheses hold karti hain, $c$ dhundho	Ex 1
B. Plain Rolle	$f (a) = f (b)$ , flat point dhundho	Ex 2
C. Multiple $c$	andar kai valid points	Ex 3
D. Hypothesis fails — not differentiable	corner/cusp tod deta hai	Ex 4
E. Hypothesis fails — not continuous	jump/hole tod deta hai	Ex 5
F. Degenerate / limiting	$b \to a$ , endpoints merge ho jaate hain	Ex 6
G. Bounding inequality (both signs)	difference ko derivative mein badlo	Ex 7
H. Word problem (speed)	average vs instantaneous, units	Ex 8
I. Exam twist — count roots	Rolle use karo solutions ki count bound karne ke liye	Ex 9

Poore note mein Continuity aur Differentiability prerequisite hain, aur jab bhi koi hypothesis questionable ho tab Extreme Value Theorem aur Fermat's Theorem (interior extrema) pe lean karte hain.

Example 1 — Plain MVT (Cell A)

Worked example

$f (x) = x^{3}$ par $[0, 2]$ ke liye har $c$ dhundho Forecast: padhne se pehle guess karo — cube $x = 2$ ke paas steeply badhta hai, toh kya $c$ , $2$ ke zyada kareeb hoga ya middle ke?

Step 1 — Secant slope compute karo (target). $\frac{f ( 2 ) - f ( 0 )}{2 - 0} = \frac{8 - 0}{2} = 4.$ Yeh step kyun? MVT kehta hai $f^{'} (c)$ ko exactly is number ke barabar hona chahiye; $c$ dhundhne se pehle target chahiye.

Step 2 — Derivative likho aur target ke barabar karo. $f^{'} (x) = 3 x^{2}$ , toh solve karo $3 c^{2} = 4$ . Yeh step kyun? $c$ woh jagah hai jahan instantaneous slope $=$ average slope.

Step 3 — Solve karo, sirf interior roots rakho. $c^{2} = \frac{4}{3} \Rightarrow c = \pm \frac{2}{3} \approx \pm 1.1547.$ Negative root $(0, 2)$ ke bahar hai, toh discard karo. $c = \frac{2}{3} \approx 1.155$ rakho. Yeh step kyun? MVT sirf open interval $(a, b)$ mein $c$ ka promise karta hai; bahar ke roots count nahi hote.

Verify: $c \approx 1.155$ , $0$ aur $2$ ke beech hai ✔, aur $3 (1.155)^{2} \approx 4.00$ ✔ — secant slope se match karta hai. Tumhara forecast: $c \approx 1.15$ midpoint $1$ se right hai, kyunki cube left side pe flatter hai.

Figure s01 mein white curve $y = x^{3}$ dikhti hai $[0, 2]$ par, pale-yellow dashed secant $(0, 0)$ se $(2, 8)$ tak (slope $4$ ), aur pink tangent $c = 2/ 3 \approx 1.155$ par. Dekho pink tangent yellow secant ke exactly parallel hai, aur pink touch-point midpoint $x = 1$ ke right mein hai — forecast ka visual proof.

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

Example 2 — Plain Rolle (Cell B)

Example 3 — Multiple $c$ (Cell C)

Figure s02 mein white sine wave $[0, 2 π]$ par hai, yellow dashed horizontal secant (slope $0$ , kyunki dono endpoints $x$ -axis par hain), aur do short flat tangents: peak $c = \frac{π}{2}$ par pink aur trough $c = \frac{3 π}{2}$ par blue. Dono horizontal hain — do alag points jahan MVT ka promise poora hota hai.

Example 4 — Hypothesis fails: not differentiable (Cell D)

Worked example

$f (x) = ∣ x - 1∣$ on $[0, 2]$ — kya Rolle apply hoti hai? Forecast: endpoints equal lagte hain. Kya flat tangent hogi?

Step 1 — Endpoints check karo. $f (0) = ∣0 - 1∣ = 1$ aur $f (2) = ∣2 - 1∣ = 1$ , toh $f (a) = f (b)$ ✔. Continuous ✔. Yeh step kyun? Teen mein se do hypotheses pass ho gayi — temptation hoti hai jeet maan lo.

Step 2 — Corner $x = 1$ par differentiability check karo. $1$ ke left pe slope $- 1$ hai; right pe $+ 1$ hai. Do one-sided slopes agree nahi karte, toh $f^{'} (1)$ exist nahi karta. Yeh step kyun? Rolle $(0, 2)$ ke sab points par differentiability maangti hai; ek bhi corner guarantee void kar deta hai.

Step 3 — Confirm karo ki koi flat tangent exist nahi karti. $x = 1$ ke alaawa har jagah, $f^{'} (x) = \pm 1 \neq = 0$ . Koi $c$ nahi hai jahan $f^{'} (c) = 0$ ho. Yeh step kyun? Dikhata hai ki theorem ka conclusion genuinely fail hota hai — hypotheses decoration nahi hain.

Verify: slopes $\pm 1$ hain, kabhi $0$ nahi ✔. Yeh parent note ki $∣ x ∣$ warning hai, shift ki hui. Fix: hamesha "spiky" point test karo.

Figure s03 mein $∣ x - 1∣$ ki white V-shape dikhti hai, yellow dashed line do equal-height endpoints $(0, 1)$ aur $(2, 1)$ ko join karti hai, aur sharp corner $(1, 0)$ par pink marker hai. Annotation batata hai ki slope abruptly $- 1$ se $+ 1$ ho jaati hai — koi smooth in-between value $0$ nahi hai, toh koi horizontal tangent kahan bhi exist nahi karti.

Example 5 — Hypothesis fails: not continuous (Cell E)

Worked example

$f (x) = {x 0 0 \leq x < 1 x = 1$ on $[0, 1]$ Forecast: kya average slope achievable hai?

Step 1 — Endpoints try karo. $f (0) = 0$ , $f (1) = 0$ , toh $f (a) = f (b)$ — Rolle ki teesri condition theek lagti hai. Yeh step kyun? Yeh phir se Rolle apply karne ke liye bait karta hai.

Step 2 — $x = 1$ par continuity check karo. Jab $x \to 1^{-}$ , $f (x) \to 1$ , lekin $f (1) = 0$ . Ek jump hai — $f$ , $[0, 1]$ par continuous nahi hai. Yeh step kyun? Closed interval par continuity hypothesis 1 hai; ek jump isko khatam kar deta hai.

Step 3 — Confirm karo ki koi flat tangent nahi hai. $[0, 1)$ par slope $f^{'} (x) = 1$ har jagah hai; kahi bhi $0$ nahi hai. Yeh step kyun? Dikhata hai ki sirf non-differentiability nahi, discontinuity bhi conclusion tod deti hai.

Verify: smooth part par slope constantly $1 \neq = 0$ hai ✔. Cell D aur Cell E dono prove karte hain: teeno hypotheses hamesha check karo.

Example 6 — Degenerate / limiting case (Cell F)

Worked example Jab

$b \to a$ , MVT slope kya ban jaati hai? Forecast: interval ko ek point tak shrink karo — "average slope" kya ban jaata hai?

Step 1 — $f (x) = x^{2}$ ke liye $[a, a + h]$ par secant slope likho jahan $h > 0$ . Hum $h > 0$ lete hain taaki interval $[a, a + h]$ ke endpoints sahi order mein hon (left $= a$ , right $= a + h$ ); hum $h \to 0^{+}$ lete jaayenge taaki right se collapse ho. $\frac{f ( a + h ) - f ( a )}{h} = \frac{( a + h ) ^{2} - a ^{2}}{h} = \frac{2 ah + h ^{2}}{h} = 2 a + h .$ Yeh step kyun? Hum dekhna chahte hain ki MVT kya report karta hai jab interval collapse ho ( $h \to 0^{+}$ ); $h$ ka sign fix karna interval well-defined rakhta hai.

Step 2 — MVT jo $c$ deta hai woh dhundho. Set karo $f^{'} (c) = 2 c = 2 a + h \Rightarrow c = a + \frac{h}{2}$ — tiny interval ka midpoint. Yeh step kyun? $x^{2}$ ke liye guaranteed point hamesha midpoint hota hai; dekhte hain yeh kahan jaata hai.

Step 3 — $h \to 0^{+}$ ka limit lo. $c = a + \frac{h}{2} \to a$ , aur secant slope $2 a + h \to 2 a = f^{'} (a)$ . Yeh step kyun? MVT statement derivative ki definition mein degenerate ho jaata hai — average slope uss single surviving point par instantaneous slope ban jaata hai.

Verify: $a = 3$ par, midpoint slope $2 (3) = 6$ , aur $f^{'} (3) = 2 \cdot 3 = 6$ ✔. Degenerate MVT $=$ derivative. Yeh Taylor's Theorem ka seed hai.

Example 7 — Bounding inequality, both signs (Cell G)

Worked example Prove karo ki

$ln (1 + x) \leq x$ for all $x > - 1$ Forecast: MVT ek difference ko ek derivative value mein badalta hai — hum kaunse do points compare karte hain, aur kya $x$ ka sign argument badal deta hai?

Step 1 — Case $x = 0$ . $ln (1 + 0) = 0 = x$ , toh equality hold karti hai. Is case ke liye done. Yeh step kyun? Boundary case woh hai jahan equality achieve hoti hai; pehle use isolate karo.

Step 2 — Case $x > 0$ : $[0, x]$ par MVT apply karo. Maano $f (t) = ln (1 + t)$ . MVT deta hai $\frac{l n ( 1 + x ) - l n 1}{x - 0} = f^{'} (c) = \frac{1}{1 + c}, c \in (0, x) .$ Kyunki $c > 0$ , $1 + c > 1$ , toh $\frac{1}{1 + c} < 1$ . Positive $x$ se multiply karo: $ln (1 + x) < x .$ Yeh step kyun? $c > 0$ ke liye derivative $1$ se neeche hai; positive $x$ se multiply karne par inequality direction preserve hoti hai.

Step 3 — Case $- 1 < x < 0$ : $[x, 0]$ par MVT apply karo. Ab $x < 0$ hai, toh interval $[x, 0]$ hai (left $= x$ , right $= 0$ ). MVT deta hai $\frac{l n 1 - l n ( 1 + x )}{0 - x} = \frac{1}{1 + c}, c \in (x, 0) .$ Kyunki $- 1 < c < 0$ , $0 < 1 + c < 1$ hai, toh $\frac{1}{1 + c} > 1$ . Left side $\frac{- ln ( 1 + x )}{- x}$ hai; isse $> 1$ set karo: $\frac{- l n ( 1 + x )}{- x} > 1.$ Yahan $- x > 0$ hai, toh $- x$ se multiply karo: $- ln (1 + x) > - x$ , yaani $ln (1 + x) < x$ . Yeh step kyun? Negative side par derivative $1$ se zyada hai, aur run $- x$ positive hai — humari pehli reasoning ko cleanly reverse karke hum usi inequality $ln (1 + x) < x$ par pahunchte hain.

Verify: $x = 1$ : $ln 2 \approx 0.693 \leq 1$ ✔. $x = 0.5$ : $ln 1.5 \approx 0.405 \leq 0.5$ ✔. $x = - 0.5$ : $ln 0.5 \approx - 0.693 \leq - 0.5$ ✔ (negative side bhi kaam karta hai). Parent note ke $sin$ /Lipschitz Continuity example jaisi MVT technique, ab $x$ ke dono signs cover karte hue.

Example 8 — Word problem: average vs instantaneous (Cell H)

Worked example Speed camera

Ek car camera $A$ se guzarti hai, phir straight road par $12 km$ aage camera $B$ se, $6$ minutes mein. Speed limit $100 km/h$ hai. Prove karo ki driver ne zaroor kisi instant mein limit tod di. Forecast: average speed kya thi — aur MVT speedometer ke baare mein kya promise karta hai?

Step 1 — Notation set up karo aur clean units mein average speed compute karo. Maano $t_{A}$ woh clock time hai jab car camera $A$ se guzarti hai, aur $t_{B}$ woh time hai jab camera $B$ se; toh $t_{B} - t_{A} = 6$ min $= 0.1$ h. Maano $s (t)$ time $t$ par car ki road par position (km) hai, toh $s (t_{B}) - s (t_{A}) = 12$ km. Average speed $\frac{s ( t _{B} ) - s ( t _{A} )}{t _{B} - t _{A}} = \frac{12 km}{0.1 h} = 120 km/h .$ Yeh step kyun? $t_{A}, t_{B}, s (t)$ pehle define karna matlab hai ki MVT quotient mein har symbol earned hai; yeh average exactly position-vs-time ka secant slope hai.

Step 2 — MVT se model karo. $s (t)$ continuous aur differentiable hai (real motion smooth hoti hai). MVT ek time $c \in (t_{A}, t_{B})$ deta hai jahan $s^{'} (c) = \frac{s ( t _{B} ) - s ( t _{A} )}{t _{B} - t _{A}} = 120 km/h .$ Yeh step kyun? $s^{'} (c)$ instantaneous speedometer reading hai — MVT force karta hai ki yeh average ke barabar kahi na kahi ho.

Step 3 — Conclude karo. Us instant $c$ par, speed exactly $120 > 100$ km/h thi — limit se upar. Guilty. Yeh step kyun? Theorem ek average fact (easy to measure) ko ek instantaneous fact (offense) mein convert karta hai.

Verify: units: $\frac{km}{h}$ ✔; $120$ km/h $> 100$ km/h ✔. Yeh parent note ki "100 km/h speedometer" story rigorous bana di.

Example 9 — Exam twist: counting roots (Cell I)

Worked example Dikhao ki

$p (x) = x^{3} + 3 x + 1$ ka exactly ek real root hai Forecast: ek cubic mein 1 ya 3 real roots ho sakti hain. Rolle extras rule out kar sakti hai — kaise?

Step 1 — Kam se kam ek root ka existence. $p (- 1) = - 1 - 3 + 1 = - 3 < 0$ aur $p (0) = 1 > 0$ . Continuity se, $p$ , $- 1$ aur $0$ ke beech zero cross karta hai — at least ek real root. Yeh step kyun? Intermediate-Value idea (Continuity ka consequence) ek crossing guarantee karta hai.

Step 2 — Maano do roots hain, aur Rolle invoke karo. Agar $p$ ke do roots $r_{1} < r_{2}$ hote, toh $p (r_{1}) = p (r_{2}) = 0$ , toh Rolle ek $c \in (r_{1}, r_{2})$ deti hai jahan $p^{'} (c) = 0$ . Yeh step kyun? Do equal-height points (dono zero) Rolle trigger karte hain — unke beech ek flat tangent honi chahiye.

Step 3 — Dikhao ki $p^{'}$ kabhi zero nahi hoti, contradiction force karo. $p^{'} (x) = 3 x^{2} + 3 = 3 (x^{2} + 1) \geq 3 > 0$ har $x$ ke liye. Toh $p^{'} (c) = 0$ impossible hai. Yeh step kyun? Koi flat point exist nahi kar sakta ⇒ do roots impossible hain ⇒ at most ek root. Step 1 ke saath combine karo: exactly ek.

Verify: $p^{'} (x) = 3 x^{2} + 3 > 0$ hamesha ✔; $p (- 1) = - 3$ , $p (0) = 1$ opposite signs ✔. (Rolle "no second root" filter ki tarah use hui — ek classic Increasing and Decreasing Functions argument disguise mein.)

Recall Quick self-test

$∣ x - 1∣$ ke liye $[0, 2]$ par Rolle fail hoti hai kyunki... ::: yeh corner $x = 1$ par differentiable nahi hai. $x^{3}$ ke liye $[0, 2]$ par MVT $c$ hai... ::: $c = \frac{2}{3} \approx 1.155$ . Jab $b \to a$ , MVT statement ban jaati hai... ::: derivative ki definition $f^{'} (a)$ . Rolle se cubic ka sirf ek real root kaise prove karte hain? ::: maano do roots hain, unke beech $p^{'} (c) = 0$ milta hai, phir dikhao $p^{'} > 0$ har jagah — contradiction. Average speed $120$ km/h, $100$ km/h zone mein speeding prove karta hai kyunki... ::: MVT force karta hai ki instantaneous speed kisi instant par exactly $120$ km/h ke barabar ho.

Connections

Extreme Value Theorem — woh max/min jis par Rolle khadi hai.
Fermat's Theorem (interior extrema) — kyun interior peak par $f^{'} = 0$ hota hai.
Continuity · Differentiability — do hypotheses jo Cells D aur E mein test hoti hain.
Increasing and Decreasing Functions — $f^{'}$ ka sign jo root-counting twist mein use hota hai.
Cauchy's Mean Value Theorem · Taylor's Theorem — jahan degenerate case (Cell F) aage le jaata hai.
Lipschitz Continuity — Cell G ki bounding technique.