4.1.28
Derivative f ′ ( x ) x slope hota hai. Slope batata hai ki road kis taraf jhuki hai:
Jab right ki taraf chalte hue upar jhuke → function increasing hai → f ′ ( x ) > 0
Neeche jhuke → function decreasing hai → f ′ ( x ) < 0 Flat (f ′ ( x ) = 0
Toh f ′ sign ek complete map hai — function kahan rise aur fall karti hai. Local maxima aur minima exactly wahan hote hain jahan slope sign switch kare.
Definition Increasing / Decreasingf I increasing hai agar I a < b f ( a ) < f ( b ) f I decreasing hai agar a < b ⇒ f ( a ) > f ( b )
First principles se derivation (Mean Value Theorem).
I a < b f closed interval [ a , b ] aur open interval ( a , b ) hai. Tab Mean Value Theorem kehta hai ki koi c ∈ ( a , b ) f ′ ( c ) = b − a f ( b ) − f ( a ) .
Ye step kyun? MVT guarantee karta hai ek aisa point jahan instantaneous slope, [ a , b ]
Ab rearrange karo:
f ( b ) − f ( a ) = f ′ ( c ) ( b − a ) . b − a > 0
Agar I f ′ > 0 f ′ ( c ) > 0 f ( b ) − f ( a ) > 0 f ( b ) > f ( a ) increasing . ✓
Agar har jagah f ′ < 0 f ( b ) − f ( a ) < 0 decreasing . ✓
Yahi poora theorem hai — derive kiya, memorise nahi.
Definition Critical pointc f critical point hai agar f ′ ( c ) = 0 ya f ′ ( c ) c sirf critical points par ho sakte hain (Fermat's theorem).
Intuition Sign-change rule kyun kaam karti haiPehle upar phir neeche jaana matlab tum pahad ki choti par the → max. Pehle neeche phir upar jaana matlab tum valley ke neeche the → min. Agar slope dono taraf positive tha, toh tum bas ek momentary pause par ruke (jaise ek seedhi chadhai par ek chhota sa flat step) — koi peak nahi.
f ′ ( x ) Critical points dhundho: f ′ ( x ) = 0 f ′
Unhe number line par mark karo, jisse intervals ban jayein.
Har interval mein ek test point lo, f ′
Increasing/decreasing intervals padho aur har critical point classify karo.
Worked example Example 1 — ek cubicf ( x ) = x 3 − 3 x
Step 1: f ′ ( x ) = 3 x 2 − 3 Kyun? Term-by-term power rule.
Step 2: 3 x 2 − 3 = 0 ⇒ x 2 = 1 ⇒ x = ± 1 − 1 , 1
Step 3–4: Intervals ( − ∞ , − 1 ) , ( − 1 , 1 ) , ( 1 , ∞ )
Test x = − 2 f ′ = 3 ( 4 ) − 3 = 9 > 0
Test x = 0 f ′ = − 3 < 0
Test x = 2 f ′ = 9 > 0
Test points kyun? f ′
Step 5: x = − 1 + → − local max , value f ( − 1 ) = − 1 + 3 = 2 x = 1 − → + local min , value f ( 1 ) = 1 − 3 = − 2
Worked example Example 2 — critical point jo extremum NAHI haif ( x ) = x 3 f ′ ( x ) = 3 x 2 = 0 x = 0 x = − 1 f ′ = 3 > 0 x = 1 f ′ = 3 > 0 + → + no extremum (ye horizontal tangent wala inflection hai).
Ye kyun important hai: f ′ ( c ) = 0
Worked example Example 3 — derivative undefined (corner)f ( x ) = ∣ x ∣ x < 0 f ′ ( x ) = − 1 x > 0 + 1 f ′ ( 0 ) x = 0 − → + local (aur global) minimum , bhale hi f ′ ( 0 ) Ye step kyun? Critical points mein "derivative undefined" bhi shaamil hai, isliye test kink par bhi kaam karta hai.
Common mistake Steel-manned common errorsMistake A: "f ′ ( c ) = 0 c Kyun sahi lagta hai: har smooth peak/valley par f ′ = 0 Fix: x 3 f ′ ( 0 ) = 0
Mistake B: Jahan f ′ Kyun sahi lagta hai: hum usually "f ′ = 0 Fix: corners/cusps/vertical tangents (jaise ∣ x ∣ x 2/3
Mistake C: Test ke liye critical point khud ko f ′ Kyun sahi lagta hai: wahi value hai jis par dhyan hai. Lekin f ′ ( c ) = 0 Fix: c left aur right mein test points lo.
Mistake D: Critical x value f ( c ) c
Recall Active recall — answers chhupa lo
Increasing ke liye f ′ f ′ ( x ) > 0
Sign rule prove karne wala theorem kaun sa hai? ::: Mean Value Theorem
Point critical hone ke do tarike? ::: f ′ ( c ) = 0 f ′ ( c )
+ → − Kya f ′ ( c ) = 0 x 3
MVT ke hypotheses? ::: f [ a , b ] ( a , b )
Recall Feynman: ek 12-saal ke bacche ko explain karoSocho tum graph ke saath bike chala rahe ho. Jab road upar chadh rahi ho, function bada ho raha hai (slope positive). Neeche utarte waqt, chhota ho raha hai (slope negative). Pahad ki choti par ya valley ke neeche bilkul flat spots hain jahan slope zero hai. Yeh jaanne ke liye ki flat spot hilltop hai ya valley, bas pehle aur baad mein dekho: pehle upar gaye phir neeche = hilltop (max); pehle neeche gaye phir upar = valley (min). Agar dono taraf upar chadh rahe the, toh tum bas ek chhote se flat bump se guzre — koi real peak nahi.
Mnemonic Test yaad karne ka trick"PND for a Mountain, NPD for a Valley."
P ositive→N egative→D ip down = peak (Max ).
N egative→P ositive = bounce up = trough (Min ).
Same sign = "S ame, S o nothing" (no extremum).
Define an increasing function on an interval I I a < b f ( a ) < f ( b )
State the link between f ′ f ′ > 0 I f ′ < 0
State the hypotheses of the Mean Value Theorem. f [ a , b ] ( a , b )
Which theorem justifies the monotonicity rule, and how? MVT: f ( b ) − f ( a ) = f ′ ( c ) ( b − a ) b − a > 0 f ′ ( c ) f ( b ) − f ( a )
Define a critical point. Domain ka woh point jahan f ′ ( c ) = 0 f ′ ( c )
State the First Derivative Test. Critical c f ′ + → − − → +
Why isn't f ′ ( c ) = 0 Slope bina sign switch kiye zero ho sakti hai (e.g. x 3 0
Classify extrema of f = x 3 − 3 x Local max at x = − 1 x = 1 − 2
Why do we use test points between critical points? f ′
Give a critical point where f ′ f = ∣ x ∣ x = 0
Mean Value Theorem — sign rule ka engine.Fermat's Theorem on Stationary Points — extrema sirf critical points par hote hain.Second Derivative Test — concavity f ′′ Concavity and Inflection Points — f ′′ f ′ Optimization (Closed Interval Method) — ise global extrema tak extend karta hai.Curve Sketching — f ′ f ′′
sign changes plus to minus
sign changes minus to plus
number line and test points
