Mean Value Theorem — proof, Rolle's theorem: Exercises

4.1.27 · D4 · Maths › Calculus I — Limits & Derivatives › Mean Value Theorem — proof, Rolle's theorem

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Level 1 — Recognition

Exercise 1.1

$f (x) = x^{2} - 6 x + 8$ ke liye $[2, 4]$ par, check karo ki Rolle ke teeno hypotheses hold karte hain ya nahi. $c$ abhi mat dhundho — bas har condition verify (ya reject) karo.

Recall Solution

"Hypotheses check karna" ka matlab: teen requirements ki list mein se har ek ko tick karo. Solve karne se pehle kyun bother karein? Rolle flat tangent ka promise tabhi karta hai jab teeno hold karein; check skip karo aur jo bhi $c$ tum compute karo woh algebra ka fluke ho sakta hai, theorem ki guarantee nahi.

$[2, 4]$ par continuous? $f$ ek polynomial hai. Polynomials har jagah continuous hote hain, toh haan. ✔ (Dekho Continuity.)
$(2, 4)$ par differentiable? Polynomials har jagah differentiable hote hain, toh haan. ✔ (Dekho Differentiability.)
Equal endpoints, $f (2) = f (4)$ ? $f (2) = 4 - 12 + 8 = 0, f (4) = 16 - 24 + 8 = 0.$ Dono $0$ ke barabar hain, toh $f (2) = f (4)$ . ✔ Teeno hold karte hain — Rolle $(2, 4)$ ke andar flat tangent ki guarantee deta hai, bhale hi humne abhi usse locate nahi kiya.

Exercise 1.2

$g (x) = x$ ke liye $[0, 9]$ par, woh secant slope identify karo jo MVT ke $c$ ko match karni chahiye. (Bas woh number compute karo jo theorem promise karta hai ki derivative ushe equal karegi.)

Recall Solution

MVT ka target number secant slope $\frac{g ( b ) - g ( a )}{b - a}$ hai — total rise over total run. Ye number kyun? MVT kehta hai ki derivative kahin average rate of change ke barabar hogi; woh average rate exactly yahi secant slope hai, toh isse compute karne se hum pehle se jaante hain ki $g^{'} (c)$ kaunsi value hit karne ka promise hai. $\frac{g ( 9 ) - g ( 0 )}{9 - 0} = \frac{9 - 0}{9} = \frac{3 - 0}{9} = \frac{1}{3} .$ Toh MVT promise karta hai ki koi $c \in (0, 9)$ hoga jahan $g^{'} (c) = \frac{1}{3}$ . (Note karo ki $g$ , $[0, 9]$ par continuous hai aur open interval $(0, 9)$ par differentiable — derivative $\frac{1}{2 x}$ , $x = 0$ par blow up karti hai, lekin $0$ ek endpoint hai, wahan differentiable hona zaroori nahi.)

Level 2 — Application

Exercise 2.1

$f (x) = x^{2}$ , $[- 1, 2]$ par. MVT se guaranteed har $c \in (- 1, 2)$ dhundho.

Recall Solution

Step 1 — target (secant slope). Kyun: MVT kehta hai $f^{'} (c)$ , $[- 1, 2]$ par average rate of change ke barabar hoga, toh pehle hum woh average compute karte hain — yahi woh number hai jo derivative ko hit karni hai. $\frac{f ( 2 ) - f ( - 1 )}{2 - ( - 1 )} = \frac{4 - 1}{3} = 1.$ Step 2 — derivative barabar rakho. $f^{'} (c) =$ secant slope kyun rakhen? Kyunki woh equation literally MVT conclusion hai; use solve karne se promised point milta hai. $f^{'} (x) = 2 x$ , toh $2 c = 1$ solve karo: $c = \frac{1}{2} .$ Step 3 — check karo ki andar hai. Kyun: MVT sirf open interval mein $c$ ki guarantee deta hai, toh valid answer ko $c \in (- 1, 2)$ satisfy karna chahiye. Kya $\frac{1}{2} \in (- 1, 2)$ ? Haan. ✔ Yahan exactly ek $c$ hai.

Exercise 2.2

$f (x) = x^{3} - x$ , $[- 1, 1]$ par. Rolle apply hota hai (jaldi check karo). Saare points $c$ dhundho jahan $f^{'} (c) = 0$ ho.

Recall Solution

Hypotheses: polynomial ⇒ continuous aur differentiable; $f (- 1) = - 1 + 1 = 0$ , $f (1) = 1 - 1 = 0$ , toh $f (- 1) = f (1)$ . ✔ Rolle apply hota hai. $c$ dhundho: Yahan $f^{'} (c) = 0$ kyun set karein (secant slope nahi)? Rolle, MVT ka special case hai equal endpoints ke saath, toh secant slope $0$ hai; promised point woh hai jahan tangent flat ho, yani $f^{'} (c) = 0$ . $f^{'} (x) = 3 x^{2} - 1$ , toh $3 c^{2} - 1 = 0 \Rightarrow c^{2} = \frac{1}{3} \Rightarrow c = \pm \frac{1}{3} \approx \pm 0.577.$ Dono $+ \frac{1}{3}$ aur $- \frac{1}{3}$ , $(- 1, 1)$ mein hain. Ye "at least one, maybe more" rule ka live example hai — yahan do flat tangents hain: cubic $x \approx - 0.577$ ke paas valley mein jaata hai aur $x \approx + 0.577$ ke paas hump tak uthta hai, aur har turning point par tangent horizontal hai, jo equal-height endpoints ko join karne wale slope- $0$ secant se match karta hai.

Exercise 2.3

$f (x) = ln x$ , $[1, e]$ par. MVT se $c$ dhundho.

Recall Solution

Step 1 — secant slope. Natural log ke endpoints kyun: $ln 1 = 0$ , $ln e = 1$ , aur yahi average rate woh value hai jo MVT $f^{'} (c)$ ko equal karne par force karti hai. $\frac{l n e - l n 1}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} .$ Step 2 — derivative. $\frac{1}{c}$ use karke equal kyun set karein? $f^{'} (c) =$ secant slope solve karna $ln$ par apply MVT conclusion hai. $f^{'} (x) = \frac{1}{x}$ , toh $\frac{1}{c} = \frac{1}{e - 1}$ deta hai $c = e - 1 \approx 1.718.$ Step 3 — andar hai? Kyun: sirf interior $c$ count hota hai. $1 < e - 1 < e$ kyunki $e - 1 \approx 1.718$ aur $e \approx 2.718$ . ✔

Level 3 — Analysis

Exercise 3.1

Kya Rolle's theorem $f (x) = ∣ x - 2∣$ par $[0, 4]$ par apply hota hai? Agar $f^{'} (c) = 0$ wala $c$ exist karta hai toh use dhundho; agar nahi, toh precisely batao kaun sa hypothesis fail hota hai aur kyun.

Recall Solution

Endpoints: $f (0) = ∣0 - 2∣ = 2$ , $f (4) = ∣4 - 2∣ = 2$ . Equal. ✔ Continuity: absolute value har jagah continuous hai. ✔ $(0, 4)$ par Differentiability? Ye decisive check kyun hai: Rolle ki guarantee usi instant void ho jaati hai jab koi bhi ek hypothesis toot jaaye, toh hum break dhundhte hain. Graph ek "V" hai jiska corner $x = 2 \in (0, 4)$ par hai. Corner par left slope $- 1$ hai aur right slope $+ 1$ — ye agree nahi karte, toh $f^{'} (2)$ exist nahi karta. Hypothesis 2 fail hoti hai. Conclusion: Rolle apply nahi hota, aur waakai $f^{'} (x) = \pm 1$ kabhi $0$ ke barabar nahi hota. Aisa koi $c$ exist nahi karta. Moral: baaki saare boxes tick hona kaafi nahi — ek broken hypothesis guarantee void kar deti hai.

Exercise 3.2

$f (x) = \frac{1}{x}$ , $[- 1, 1]$ par. Koi likhta hai: " $\frac{f ( 1 ) - f ( - 1 )}{1 - ( - 1 )} = \frac{1 - ( - 1 )}{2} = 1$ , aur $f^{'} (x) = - \frac{1}{x ^{2}}$ , toh $- \frac{1}{c ^{2}} = 1$ deta hai $c^{2} = - 1$ : koi real $c$ nahi! MVT toot gayi." Explain karo ki asal mein kya galat hua.

Recall Solution

Flaw MVT mein nahi hai — actually yeh hai ki MVT ko run karne ki permission hi nahi mili. Check kyun matter karta hai: MVT ka conclusion tabhi promise hota hai jab uske hypotheses hold karein, toh kisi bhi equation par trust karne se pehle hum unhe verify karte hain. MVT demand karta hai ki $f$ poore closed interval $[- 1, 1]$ par continuous ho. Lekin $\frac{1}{x}$ ka vertical asymptote $x = 0$ par hai, jo $[- 1, 1]$ ke andar hai: function wahan defined hi nahi, continuous toh bilkul bhi nahi. Toh hypothesis 1 fail hoti hai aur theorem koi promise nahi karta. "Contradiction" $c^{2} = - 1$ exactly wahi hai jo tab expect karoge jab tum theorem ko uske domain ke bahar apply karo — algebra sahi report karti hai ki koi interior tangent slope $1$ nahi hai, kyunki graph gap ke across $- \infty$ se $+ \infty$ tak jump karta hai. Fix: pehle $[a, b]$ ko scan karo kisi bhi aisi point ke liye jahan $f$ undefined ya discontinuous ho. Agar koi ho, ruk jao — MVT inapplicable hai.

Exercise 3.3

Maano $f$ , $R$ par differentiable hai aur har $x$ ke liye $f^{'} (x) = 0$ hai. MVT use karke prove karo ki $f$ constant hai.

Recall Solution

Goal: dikhao ki kisi bhi do inputs $x < y$ ke liye $f (x) = f (y)$ hai, jo exactly "constant" ka matlab hai. MVT apply karo $[x, y]$ par (har jagah continuous + differentiable, toh hypotheses hold karte hain). Yahan MVT kyun? Yahi ek tool hai jo difference $f (y) - f (x)$ ko ek single derivative value mein convert karta hai jise hum control kar sakte hain. Koi $c \in (x, y)$ hai jahan $f^{'} (c) = \frac{f ( y ) - f ( x )}{y - x} .$ Lekin assumption ke anusaar $f^{'} (c) = 0$ . Ye kyun finish karta hai: left side $0$ par force hai, toh right side (average change) bhi $0$ hai, matlab kisi bhi do points ke beech koi net change nahi. Toh $\frac{f ( y ) - f ( x )}{y - x} = 0 \Rightarrow f (y) - f (x) = 0 \Rightarrow f (y) = f (x) .$ Kyunki $x, y$ arbitrary the, $f$ har jagah same value leta hai — woh constant hai. $■$ Yahi "zero derivative ⇒ flat function" ki backbone hai, jo Increasing and Decreasing Functions ke peeche ka tool hai.

Level 4 — Synthesis

Exercise 4.1

Dikhao ki equation $x^{3} + 3 x - 1 = 0$ ka exactly one real root hai. (Hint: existence continuity se, uniqueness Rolle se.)

Recall Solution

Maano $f (x) = x^{3} + 3 x - 1$ . Existence (at least one root) — Intermediate Value Theorem. Intermediate Value Theorem (IVT) kehta hai: agar $f$ , $[a, b]$ par continuous hai aur endpoints par opposite sign ki values leta hai, toh $f$ beech ki har value hit karta hai — khaaskar $0$ ko woh $(a, b)$ mein kahin equal karta hi hai. Humen iska kyun chahiye: yeh sign change se ek root manufacture karta hai. Uske hypotheses check karo: $f$ ek polynomial hai, toh $[0, 1]$ par continuous ✔, aur $f (0) = - 1 < 0, f (1) = 1 + 3 - 1 = 3 > 0,$ opposite signs ✔. Isliye IVT $(0, 1)$ mein at least one root deta hai. Uniqueness (no second root) — contradiction ke saath Rolle use karke. Rolle kyun: do roots do equal-height points honge ( $f = 0$ dono par), exactly Rolle ka setup, jo ek flat tangent force karega jise hum phir rule out kar sakte hain. Maano do roots $a < b$ hote, toh $f (a) = f (b) = 0$ . Phir $f$ , $[a, b]$ par continuous, $(a, b)$ par differentiable, aur equal endpoint values wala hai — Rolle apply hota hai, aur koi $c$ milta hai jahan $f^{'} (c) = 0$ ho. Lekin $f^{'} (x) = 3 x^{2} + 3 = 3 (x^{2} + 1) \geq 3 > 0 for all x .$ Derivative kabhi zero nahi hoti, jo $f^{'} (c) = 0$ se contradict karta hai. Toh do roots impossible hai. Conclusion: exactly one real root. $■$

Exercise 4.2

Prove karo ki saare real $a, b$ ke liye, $∣ arctan b - arctan a ∣ \leq ∣ b - a ∣$ . (Yeh dikhata hai ki $arctan$ , Lipschitz hai constant $1$ ke saath.)

Recall Solution

Idea: MVT function values ka ek difference ek single derivative value mein convert karta hai jise hum bound kar sakte hain. Yeh kyun help karta hai: $arctan b - arctan a$ jaisa difference directly estimate karna mushkil hai, lekin ek derivative value $\frac{1}{1 + c ^{2}}$ bound karna aasaan hai. Maano $f (x) = arctan x$ , saare $R$ par continuous aur differentiable. Case $a \neq = b$ : $a, b$ endpoints wale interval par MVT apply karo. Koi $c$ unke beech hai jahan $\frac{a r c t a n b - a r c t a n a}{b - a} = f^{'} (c) = \frac{1}{1 + c ^{2}} .$ Absolute values lo: $\frac{∣ a r c t a n b - a r c t a n a ∣}{∣ b - a ∣} = \frac{1}{1 + c ^{2}} .$ Bound ab kyun nikal aata hai: $c^{2} \geq 0$ , toh $1 + c^{2} \geq 1$ , isliye $\frac{1}{1 + c ^{2}} \leq 1$ . Isliye $∣ arctan b - arctan a ∣ \leq ∣ b - a ∣.$ Case $a = b$ : dono sides $0$ hain, toh inequality trivially hold karti hai. $■$

Exercise 4.3

Maano $f$ , $[0, 4]$ par differentiable hai, $f (0) = 1$ hai, aur saare $x$ ke liye $f^{'} (x) \leq 3$ hai. $f (4)$ ki largest possible value kya hai? MVT se justify karo.

Recall Solution

MVT $[0, 4]$ par apply karo: MVT kyun: yeh unknown endpoint value $f (4)$ ko ek derivative value se link karta hai jis par humara ceiling hai. Koi $c \in (0, 4)$ hai jahan $f^{'} (c) = \frac{f ( 4 ) - f ( 0 )}{4 - 0} = \frac{f ( 4 ) - 1}{4} .$ Bound use karo. Yeh maximum kyun pin down karta hai: kyunki har derivative value $\leq 3$ hai, khaaskar $f^{'} (c) \leq 3$ , jo average change aur isliye $f (4)$ ko cap karta hai: $\frac{f ( 4 ) - 1}{4} \leq 3 \Rightarrow f (4) - 1 \leq 12 \Rightarrow f (4) \leq 13.$ Largest possible value: $f (4) = 13$ , straight line $f (x) = 1 + 3 x$ se achieve hota hai (jiska $f^{'} \equiv 3$ hai, har jagah bound hit karta hai). $■$

Level 5 — Mastery

Exercise 5.1 (Cauchy's MVT)

Cauchy's Mean Value Theorem kehta hai: agar $f$ aur $g$ $[a, b]$ par continuous hain, $(a, b)$ par differentiable hain, aur saare $x \in (a, b)$ ke liye $g^{'} (x) \neq = 0$ hai, toh koi $c \in (a, b)$ exist karta hai jahan $\frac{f ^{'} ( c )}{g ^{'} ( c )} = \frac{f ( b ) - f ( a )}{g ( b ) - g ( a )} .$ (Condition $g^{'} \neq = 0$ on $(a, b)$ yeh bhi guarantee karta hai ki $g (b) - g (a) \neq = 0$ , toh right side well-defined hai.) $f (x) = x^{2}$ , $g (x) = x^{3}$ , $[1, 2]$ par lo aur $c$ dhundho.

Recall Solution

Step 0 — hypotheses check karo, khaaskar $g^{'} \neq = 0$ on $(1, 2)$ . Kyun: Cauchy's MVT require karta hai ki $g^{'} (x) \neq = 0$ poore open interval mein ho, warna ratio $\frac{f ^{'} ( c )}{g ^{'} ( c )}$ undefined ho sakti hai aur $g (2) - g (1)$ $0$ ho sakta hai. $f, g$ dono polynomials hain, toh $[1, 2]$ par continuous aur $(1, 2)$ par differentiable ✔. Aur $g^{'} (x) = 3 x^{2}$ , toh $x \in (1, 2)$ ke liye $3 x^{2} \geq 3 > 0$ : $g^{'}$ kabhi vanish nahi karta. ✔ (Consistently, $g (2) - g (1) = 8 - 1 = 7 \neq = 0$ .) Cauchy's MVT ab licensed hai. Step 1 — right-hand side (total changes ka ratio). Kyun: yahi woh target value hai jo derivative-ratio ko equal karni chahiye. $\frac{f ( 2 ) - f ( 1 )}{g ( 2 ) - g ( 1 )} = \frac{4 - 1}{8 - 1} = \frac{3}{7} .$ Step 2 — left-hand side (derivatives ka ratio). Pehle differentiate kyun karein: Cauchy ke conclusion ka left side $f^{'} (c) / g^{'} (c)$ hai, derivatives ka ratio, original values ka nahi. $f^{'} (x) = 2 x$ , $g^{'} (x) = 3 x^{2}$ , toh $\frac{f ^{'} ( c )}{g ^{'} ( c )} = \frac{2 c}{3 c ^{2}} = \frac{2}{3 c} .$ Step 3 — equal set karo aur solve karo. Kyun: dono sides ko equate karna Cauchy conclusion hai; use solve karna $c$ locate karta hai. $\frac{2}{3 c} = \frac{3}{7} \Rightarrow 14 = 9 c \Rightarrow c = \frac{14}{9} \approx 1.556.$ Andar hai? $1 < \frac{14}{9} < 2$ . ✔ (Ordinary MVT, $g (x) = x$ wala special case hai, jahan $g^{'} = 1$ aur ratio plain secant slope ban jaata hai.)

Exercise 5.2 (A sharper bound via a second-order idea)

Maano $f$ twice differentiable hai, $f (0) = 0$ , $f^{'} (0) = 0$ , aur saare $x$ ke liye $∣ f^{''} (x) ∣ \leq 2$ hai. Dikhao ki $∣ f (1) ∣ \leq 1$ . (Hint: $f^{'}$ par MVT apply karo, phir $f$ se relate karo — yahi Taylor's Theorem ka seed hai.)

Recall Solution

Step 1 — slope $f^{'}$ ko $f^{''}$ ke bound se control karo. $f^{'}$ par MVT kyun apply karein: hume $f^{''}$ ( $f^{'}$ ki derivative) par ceiling pata hai, aur MVT exactly derivative bound se function tak ka bridge hai — yahan "function" $f^{'}$ hai. Kisi bhi $x \in (0, 1]$ ke liye, $f^{'}$ par $[0, x]$ par MVT apply karo: koi $t \in (0, x)$ hai jahan $f^{''} (t) = \frac{f ^{'} ( x ) - f ^{'} ( 0 )}{x - 0} = \frac{f ^{'} ( x )}{x} (using f^{'} (0) = 0) .$ Rearrange karke, $f^{'} (x) = x f^{''} (t)$ , toh $∣ f^{'} (x) ∣ = ∣ x ∣ ∣ f^{''} (t) ∣ \leq ∣ x ∣ \cdot 2 = 2∣ x ∣ = 2 x (0 < x \leq 1) .$ Yeh progress kyun hai: humne sirf $∣ f^{''} ∣ \leq 2$ jaanke shuru kiya aur ab $[0, 1]$ ke har point par $f^{'}$ ka bound hai. Step 2 — woh slope bound $[0, 1]$ par accumulate karke $f (1)$ bound karo. Ek aur MVT ki jagah integrate kyun karein: ek single MVT sirf $∣ f (1) ∣ = ∣ f^{'} (s) ∣ \leq 2 s < 2$ kisi ek point $s$ ke liye dega, jo bahut weak hai. Integrate karne se capture hota hai ki $f^{'}$ $0$ se start hoti hai aur gradually badhti hai, estimate ko aadha kar deta hai. Kyunki $f (0) = 0$ , $f (1) = f (1) - f (0) = \int_{0}^{1} f^{'} (x) d x,$ toh absolute values leke aur $∣ f^{'} (x) ∣ \leq 2 x$ use karke, $∣ f (1) ∣ = \int_{0}^{1} f^{'} (x) d x \leq \int_{0}^{1} ∣ f^{'} (x) ∣ d x \leq \int_{0}^{1} 2 x d x = [x^{2}]_{0}^{1} = 1.$ Conclusion: $∣ f (1) ∣ \leq 1$ . $■$ Extra care kyun matter karta hai: yeh "derivative bound karo, phir use interval par accumulate karo" move exactly Taylor remainder ki logic hai — tum derivative ki growth ko jitna closely dekho, function par bound utna hi sharp hoga.

Recall Poori ladder ki one-line summary

Recognition (hypotheses padho) ::: continuity, differentiability, aur — Rolle ke liye — equal endpoints check karo. Application ( $c$ dhundho) ::: $f^{'} (c)$ ko secant slope ke barabar set karo, solve karo, confirm karo ki $c \in (a, b)$ hai. Analysis (kyun fail hota hai) ::: broken hypothesis locate karo — corner, asymptote, ya jump. Synthesis (cheezein prove karo) ::: MVT differences ko single derivative values mein convert karta hai jinhein tum bound kar sako. Mastery (generalise karo) ::: Cauchy's MVT do functions ko ratio karta hai ( $g^{'} \neq = 0$ chahiye); interval par derivative bound accumulate karna Taylor's theorem ko seed karta hai.