Mean Value Theorem — proof, Rolle's theorem: Question bank

4.1.27 · D5 · Maths › Calculus I — Limits & Derivatives › Mean Value Theorem — proof, Rolle's theorem

Do roles jo straight rakhni hain:

Hypotheses = "entry ticket" conditions H1 aur H2 (aur Rolle ke liye H3). Ek bhi toodo, aur theorem kuch promise nahi karta.
Conclusion = guarantee: kam se kam ek interior point $c$ exist karta hai jahan tangent slope secant slope ke barabar ho (ya Rolle ke liye $0$ ).

Traps is hisaab se split hote hain ki tum in dono mein se kise galat handle karte ho.

Error dhundho

f (x) = tan x

[0, π]

mein

f (0) = f (π) = 0

hai, toh Rolle se koi

c

hoga jahan

f^{'} (c) = 0

Error:

tan x

x = π /2

par discontinuous hai (wahan blow up ho jaata hai), isliye ye

[0, π]

par continuous nahi hai. Hypothesis 1 (H1) fail ho jaati hai aur koi flat tangent exist nahi karta.

f (x) = x^{2/3}

[- 1, 1]

mein

f (- 1) = f (1) = 1

satisfy karta hai, toh Rolle

f^{'} (c) = 0

deta hai."

Error:

f^{'} (x) = \frac{2}{3} x^{- 1/3}

x = 0

par undefined hai (vertical tangent, ek cusp), isliye

f

poore open interval par differentiable nahi hai (H2 fail ho jaati hai). Theorem apply nahi hota.

"MVT se

f (x) = 1/ x

over

[- 1, 1]

par, ek

c

exist karta hai jahan

f^{'} (c) = \frac{f ( 1 ) - f ( - 1 )}{2} = 1

Error:

1/ x

x = 0 \in [- 1, 1]

par define hi nahi hai (continuous hona toh door ki baat), isliye H1 fail ho jaati hai. Interval domain mein hona chahiye jahan hypotheses hold karein; ek discontinuity ke upar se interval nahi le sakte.

"MVT mujhe

f (b) - f (a) = f^{'} (c) (b - a)

likhne deta hai aur generally exact

c

formula solve karne deta hai."

Error: MVT ek existence statement hai — ye

c

ka koi formula nahi batata. Specific functions ke liye tum

c

solve kar sakte ho, lekin koi universal expression nahi hai, aur kai valid

c

ho sakte hain.

f (x) = ⌊ x ⌋

(floor) integers ke beech constant hai, isliye Rolle

[0, 2]

par

f^{'} (c) = 0

ke saath lagta hai."

Error: Floor

x = 1

par

(0, 2)

ke andar jump karta hai, isliye wahan discontinuous hai — hypothesis 1 (H1) fail ho jaati hai. Jumps ke beech flat hona, unke across continuity ko nahi bachata.

"Kyunki MVT deta hai

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}

, agar average rate positive hai toh

f^{'} > 0

poore

(a, b)

par everywhere hai."

Error: MVT sirf ek point supply karta hai jahan slope average ke barabar ho. Baaki jagah derivative negative ho sakti hai; function ko sirf average par rise karna hai, monotonically nahi.

Why questions

Rolle ka proof Extreme Value Theorem invoke karke kyun shuru hota hai?

Humein ek guaranteed highest ya lowest point chahiye jahan flat tangent rakhe. EVT kehta hai ek continuous function ek closed bounded interval par actually max aur min attain karta hai — wahi "top/bottom" hai jahan hum differentiate karte hain.

Rolle ke proof ke Case B mein kam se kam ek extremum interior kyun hona chahiye?

Kyunki

f (a) = f (b)

: agar max

M

aur min

m

dono sirf endpoints par occur karte, toh

M = f (a) = f (b) = m

, jo constant force karta —

M \neq = m

ko contradict karta. Isliye ek extreme strictly andar hai.

Ek interior maximum kyun

f^{'} (c) = 0

force karta hai na ki sirf

f^{'} (c) \leq 0

Fermat's Theorem (interior extrema): right se approach karne par

f^{'} (c) \leq 0

milta hai, left se

f^{'} (c) \geq 0

. Differentiability dono one-sided limits ko barabar banata hai,

f^{'} (c)

exactly

0

tak squeeze ho jaata hai.

MVT ka proof helper

g

banane ke liye secant line subtract kyun karta hai?

Secant line ko

L (x) = f (a) + \frac{f ( b ) - f ( a )}{b - a} (x - a)

likho — woh straight line jo dono endpoints se guzarti hai. Tab

g (x) = f (x) - L (x)

us line se vertical gap hai, aur

g (a) = g (b) = 0

, isliye H3 hold karta hai aur Rolle apply ho jaata hai. Ye "graph ko flat tilt karta hai."

g (x) = f (x) - L (x)

differentiable kyun rehta hai jab

f

differentiable ho?

Yahan

L (x) = f (a) + \frac{f ( b ) - f ( a )}{b - a} (x - a)

ek straight line hai, jo everywhere differentiable hai; ek differentiable function minus ek differentiable line differentiable hoti hai. Isliye

g

exactly

f

ki smoothness inherit karta hai.

MVT immediately kyun prove karta hai ki

sin

Lipschitz hai?

MVT difference

sin b - sin a

cos c (b - a)

mein badal deta hai; kyunki

∣ cos c ∣ \leq 1

, hume milta hai

∣ sin b - sin a ∣ \leq ∣ b - a ∣

. Ek bounded derivative hamesha ek Lipschitz constant yield karta hai.

Agar

f

ke beech mein ek jump ho toh hum MVT use karke do points par

f

compare kyun nahi kar sakte?

Ek jump discontinuity hai, jo us interval par hypothesis 1 (H1) tod deta hai. Continuity ke bina "average slope" achieve nahi ho sakta — ek aisi step socho jo instantly climb kare bina koi beech ka slope liye.

Cauchy's Mean Value Theorem ordinary MVT se "zyada general" kyun hai?

Ye do functions ke changes compare karta hai,

\frac{f ( b ) - f ( a )}{g ( b ) - g ( a )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}

;

g (x) = x

set karne par ordinary MVT wapas milta hai. Ye L'Hôpital's rule aur Taylor's Theorem remainders ke peeche ka engine hai.

Edge cases

Rolle kya kehta hai agar

f

[a, b]

par constant hai?

Har hypothesis hold karta hai aur

f^{'} (x) = 0

everywhere hai, isliye koi bhi interior

c

kaam karta hai. Theorem phir bhi ek

c

deliver karta hai — bas unique nahi.

MVT ka kya hota hai jab

a = b

Interval ek point tak degenerate ho jaata hai aur secant slope

\frac{f ( b ) - f ( a )}{b - a} = \frac{0}{0}

undefined hai. Theorem ko ek genuine interval

a < b

chahiye, isliye ye case hypothesis se exclude hai.

Agar

f

[a, b]

par continuous hai aur

(a, b)

par differentiable hai lekin andar kahi ek vertical tangent (infinite slope) hai, toh kya wahan differentiable hai?

Nahi. Vertical tangent ka matlab hai derivative limit

\pm \infty

hai, finite number nahi, isliye

f

us point par differentiable nahi hai aur H2 wahan fail ho jaati hai.

Kya MVT ek unbounded interval jaise

[0, \infty)

par hold ho sakta hai?

Classical statement ko ek closed bounded interval chahiye taaki EVT (jo Rolle mein use hota hai) apply ho. Tum MVT har finite

[a, b]

par apply karte ho; endpoints finite hone chahiye.

Agar ek smooth function ki secant slope

0

hai lekin wo strictly increasing phir decreasing hai, toh

c

kahan hoga?

Turning point (peak) par, jahan

f^{'} = 0

hai. Equal endpoints kam se kam ek interior max ya min force karte hain, aur wahi flat tangent hai jo Rolle promise karta hai.

f (x) = x^{3}

[- 1, 1]

ke liye, secant slope

1

hai; kya MVT

c

exist karta hai aur kahan?

Haan.

f^{'} (x) = 3 x^{2} = 1

deta hai

c = \pm 1/ 3

, dono

(- 1, 1)

ke andar. Yahan do valid points exist karte hain — ek reminder ki

c

unique nahi hona chahiye.

Question bank — Mean Value Theorem — proof, Rolle's theorem

True ya false — justify karo

Error dhundho

Why questions

Edge cases