4.1.27
Intuition Badi picture (WHY ye matter karta hai)Agar aap 1 ghante mein 100 km drive karte ho, toh aapki average speed 100 km/h hai. MVT kehta hai: kisi ek instant par aapki instantaneous speed (speedometer reading) exactly 100 km/h thi. Ye theorem guarantee karta hai ki average slope interval ke andar kisi jagah derivative dwara achieve hoti hai. Rolle's theorem woh special case hai jahan average slope 0
Definition Rolle's TheoremMaano f
f [ a , b ] continuous hai,f ( a , b ) differentiable hai,f ( a ) = f ( b )
Toh kam se kam ek aisa point c ∈ ( a , b ) == f ′ ( c ) = 0 ==
Intuition YE kyon sach hona chahiyeAgar ek smooth curve same height par shuru aur khatam hoti hai, toh woh bina neeche aaye upar nahi ja sakti (ya bina upar aaye neeche nahi ja sakti). Turning point par — sabse oonchi peak ya sabse nichli valley par — tangent flat hoti hai, yaani f ′ ( c ) = 0 f ′ = 0
Worked example Proof — Extreme Value Theorem se banaya gayaStep 1 — Continuity use karke extremes paao.
Kyunki f [ a , b ] Extreme Value Theorem kehta hai ki f M m [ a , b ] Ye step kyon? Hume ek guaranteed "top" ya "bottom" point chahiye taaki flat tangent mil sake.
Step 2 — Do cases.
Case A: M = m Tab f constant hai, isliye f ′ ( x ) = 0 x c Kyon? Ek flat line ki slope har jagah zero hoti hai.
Case B: M = m Kyunki f ( a ) = f ( b ) M , m interior point c ∈ ( a , b ) Kyon? Agar dono extremes sirf endpoints par hote, toh M = f ( a ) = f ( b ) = m M = m
Step 3 — Interior extremum ⇒ derivative zero (Fermat's Theorem).
Maano c h > 0 f ( c + h ) ≤ f ( c ) h f ( c + h ) − f ( c ) ≤ 0 ⇒ f ′ ( c ) = lim h → 0 + ≤ 0. h < 0 f ( c + h ) ≤ f ( c ) h < 0 h f ( c + h ) − f ( c ) ≥ 0 ⇒ f ′ ( c ) = lim h → 0 − ≥ 0. f f ′ ( c ) f ′ ( c ) ≤ 0 aur f ′ ( c ) ≥ 0 f ′ ( c ) = 0 ■ Ye step kyon? Derivative dono sides se same limit hai; ise ≤ 0 ≥ 0 0
Definition Mean Value TheoremMaano f
[ a , b ] continuous , aur( a , b ) differentiable .
Toh ek aisa c ∈ ( a , b ) f ′ ( c ) = b − a f ( b ) − f ( a ) .
Intuition Geometrically YE kya kehta haiRight side b − a f ( b ) − f ( a ) secant line ki slope hai. MVT kehta hai ki andar kahin ek tangent us secant ke parallel hai.
Worked example "Graph ko tilt karo" wali trickStep 1 — Helper function banao. Define karo
g ( x ) = f ( x ) − [ f ( a ) + b − a f ( b ) − f ( a ) ( x − a ) ] . L ( x ) g ( x ) = f ( x ) − L ( x ) curve aur secant ke beech vertical gap .
Ye step kyon? Hum secant subtract karte hain taaki dono endpoints same height par aa jaayein — phir Rolle apply hota hai.
Step 2 — g
g [ a , b ] f g ( a , b ) Endpoints: g ( a ) = f ( a ) − f ( a ) = 0 g ( b ) = f ( b ) − [ f ( a ) + b − a f ( b ) − f ( a ) ( b − a ) ] = f ( b ) − f ( b ) = 0 g ( a ) = g ( b ) = 0 Ye step kyon? Secant subtract karne se endpoints same value par flatten ho jaate hain, jo Rolle ki hypothesis 3 hai.
Step 3 — g Ek aisa c ∈ ( a , b ) g ′ ( c ) = 0 g ′ ( x ) = f ′ ( x ) − b − a f ( b ) − f ( a ) . g ′ ( c ) = 0 f ′ ( c ) = b − a f ( b ) − f ( a ) . ■ Ye step kyon? Rolle hume gap function ka ek flat point deta hai; flat gap ka matlab hai curve-slope = secant-slope.
Mnemonic MVT proof yaad karo"Secant Subtract karo, Rolle ko Bulao."
Rolle bas MVT ka horizontal secant wala case hai (f ( a ) = f ( b ) ⇒ = 0
Worked example Example 1 — MVT ke liyec f ( x ) = x 2 [ 1 , 3 ] c Step 1: Secant slope = 3 − 1 f ( 3 ) − f ( 1 ) = 2 9 − 1 = 4. Kyon? Ye woh target value hai jo f ′ ( c ) Step 2: f ′ ( x ) = 2 x 2 c = 4 ⇒ c = 2. Step 3: Check karo c = 2 ∈ ( 1 , 3 ) x = 2
Worked example Example 2 — Rolle in actionf ( x ) = x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) [ 1 , 3 ] Step 1: f ( 1 ) = 0 , f ( 3 ) = 0 f ( a ) = f ( b ) Kyon? Saari Rolle hypotheses hold hoti hain ⇒ andar flat tangent guaranteed hai.
Step 2: f ′ ( x ) = 2 x − 4 = 0 ⇒ c = 2 ∈ ( 1 , 3 ) . x = 2
Worked example Example 3 — Application: function ko bound karnaDikhao ki saare a , b ∣ sin b − sin a ∣ ≤ ∣ b − a ∣ Step 1: f ( x ) = sin x c b − a sin b − sin a = cos c Step 2: Absolute values lo: ∣ b − a ∣ ∣ sin b − sin a ∣ = ∣ cos c ∣ ≤ 1. Kyon? Cosine 1 se bound hai, aur MVT difference ko ek single derivative value mein convert karta hai.
Step 3: Rearrange karo: ∣ sin b − sin a ∣ ≤ ∣ b − a ∣. ■ sin Lipschitz hai — real analysis ka ek key fact.)
Common mistake "Hypotheses optional hain — main waise bhic Kyon sahi lagta hai: Zyaadatar textbook functions nice hote hain, isliye tum check karna bhool jaate ho.
Counterexample: f ( x ) = ∣ x ∣ [ − 1 , 1 ] f ( − 1 ) = f ( 1 ) = 1 f ′ ( x ) = ± 1 0 fail hota hai kyunki f x = 0 differentiable nahi hai.
Fix: HAMESHA pehle [ a , b ] ( a , b )
Common mistake "MVT exactly ekc Kyon sahi lagta hai: Simple examples mein ek unique c Sach: Ye kam se kam ek c Fix: Ise "ek c exist karta hai " padho, na ki "ek unique c
c Kyon sahi lagta hai: Endpoints [ a , b ] Sach: c open interval ( a , b ) Fix: Hamesha c a b
Recall Feynman: ek 12-saal ke bacche ko explain karoSocho tum ek pahaad par chadhte ho aur wapas neeche uss dost ke paas aate ho jo usi height par khada hai jahan se tum chale the. Apni walk mein kahin na kahin, tum ek pal ke liye flat move kar rahe the — na upar, na neeche — pahaad ke bilkul top par. Ye Rolle's theorem hai.
Ab socho tum ek aisi jagah chalte ho jahan tumhara dost tumse zyaada upar khada hai. Tumhari "average steepness" koi number hai. MVT kehta hai ki ek pal par tumhare pair exactly usi average steepness par tilt the. Ise prove karne ke liye, hum cleverly "poora pahaad tilt karte hain" taaki end, start ki height se match kare, aur harder problem ko Rolle wale flat-hill version mein convert kar dete hain!
Rolle's Theorem ki 3 hypotheses kya hain? (1) f [ a , b ] f ( a , b ) f ( a ) = f ( b )
Rolle's Theorem ka conclusion kya hai? There exists c ∈ ( a , b ) f ′ ( c ) = 0
MVT ka conclusion batao. There exists c ∈ ( a , b ) f ′ ( c ) = b − a f ( b ) − f ( a )
MVT ka geometric matlab kya hai? Koi tangent line endpoints ko join karne wali secant ke parallel hoti hai.
Rolle, MVT ka special case kaise hai? Jab f ( a ) = f ( b ) 0 f ′ ( c ) = 0
MVT ko Rolle se prove karne wali helper function kya hai? g ( x ) = f ( x ) − [ f ( a ) + b − a f ( b ) − f ( a ) ( x − a ) ]
MVT proof mein g ( a ) = g ( b ) = 0 Kyunki g
Rolle ke proof ke liye max/min exist hone ki guarantee kaun sa theorem deta hai? The Extreme Value Theorem (continuity on a closed bounded interval).
Woh counterexample jahan non-differentiability ki wajah se Rolle fail hota hai? f ( x ) = ∣ x ∣ [ − 1 , 1 ] f ( − 1 ) = f ( 1 ) f ′ 0
Kya MVT ek unique c Nahi — kam se kam ek c
MVT use karke ∣ sin b − sin a ∣ ∣ sin b − sin a ∣ = ∣ cos c ∣ ∣ b − a ∣ ≤ ∣ b − a ∣
Extreme Value Theorem — Rolle ke proof ke liye max/min provide karta hai.Fermat's Theorem (interior extrema) — interior extremum ⇒ f ′ = 0 Continuity aur Differentiability — required hypotheses.Increasing and Decreasing Functions — MVT use karke prove hota hai (f ′ > 0 ⇒ Taylor's Theorem — MVT ko higher derivatives se generalize karta hai.Cauchy's Mean Value Theorem — do functions ke liye MVT; L'Hôpital's rule tak le jaata hai.Lipschitz Continuity — bounded derivative ⇒ Lipschitz, MVT se.
Fermat's Theorem: interior extremum
Continuous on closed interval
Differentiable on open interval
Secant slope over interval
Tangent parallel to secant
