Derivation from first principles (Mean Value Theorem).
Take any a<b in I. Assume f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then the Mean Value Theorem says there is some c∈(a,b) with
f′(c)=b−af(b)−f(a).
Why this step? MVT guarantees a point where the instantaneous slope equals the average slope over [a,b]. (Both hypotheses matter: continuity on the closed ends, differentiability inside.)
Now rearrange:
f(b)−f(a)=f′(c)(b−a).
Since b−a>0:
If f′>0 everywhere on I, then f′(c)>0, so f(b)−f(a)>0, i.e. f(b)>f(a) → increasing. ✓
If f′<0 everywhere, then f(b)−f(a)<0 → decreasing. ✓
That's the whole theorem — derived, not memorised.
Find critical points: solve f′(x)=0 and find where f′ is undefined.
Mark them on a number line, splitting it into intervals.
Pick a test point in each interval, plug into f′, record the sign.
Read off increasing/decreasing intervals and classify each critical point.
Recall Feynman: explain to a 12-year-old
Imagine riding a bike along the graph. When the road goes uphill, the function is getting bigger (slope positive). Downhill, it gets smaller (slope negative). The flat spots at the very top of a hill or bottom of a valley are where the slope is zero. To know if a flat spot is a hilltop or a valley, just look at what happens right before and right after: went up then down = hilltop (max); went down then up = valley (min). If it was uphill on both sides, you just rolled over a tiny flat bump — not a real peak.
Dekho, idea bahut simple hai: derivative f′(x) ka matlab hai graph ka slope (dhalaan). Agar slope positive hai to function upar ja raha hai — yaani increasing. Agar slope negative hai to neeche — decreasing. Aur jahan slope zero ho jaaye ya exist hi na kare, wahan critical point hota hai. Yahi se sara khel chalta hai.
Local maximum aur minimum nikalne ke liye First Derivative Test use karte hain. Critical point ke thoda left aur thoda right me f′ ka sign check karo. Agar sign + se − ho jaye (pehle chadhai, phir utraai) to wo ek pahaadi ki choti hai — local max. Agar − se + ho (pehle utraai, phir chadhai) to wo ghaati ka talla hai — local min. Agar dono taraf sign same raha, to kuch nahi — sirf ek flat moment. (Note: MVT use karne ke liye f ko [a,b] par continuous aur (a,b) par differentiable hona chahiye.)
Sabse badi galti jo students karte hain: socha ki f′(c)=0 ka matlab hamesha max/min. Galat! x3 ka f′(0)=0 hai par wahan koi extremum nahi, sirf inflection. Isliye sign change zaroor verify karo. Doosri baat — critical points sirf f′=0 wale nahi, balki jahan derivative undefined ho (jaise ∣x∣ ka corner x=0) wo bhi count hote hain.
Yeh topic kyun important hai? Kyunki real life optimization — maximum profit, minimum cost, sabse short time — sab isi se solve hota hai. Number line banao, test points lagao, sign likho, aur extrema classify kar do. Practice karoge to 30 second me ho jaayega.
