Mavzular

⭐ — imtihonda eng ko'p ishlatiladigan formulalar. · 🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

O'suvchi va kamayuvchi funksiyalar

  • dydx>0\dfrac{dy}{dx} > 0 — funksiya o'suvchi
  • dydx<0\dfrac{dy}{dx} < 0 — funksiya kamayuvchi

Barcha xx uchun dydx>0\dfrac{dy}{dx} > 0 bo'lsa, funksiya "o'suvchi funksiya" deyiladi. Masalan, y=x3+xy = x^3 + x: dydx=3x2+1>0\dfrac{dy}{dx} = 3x^2 + 1 > 0.

Statsionar nuqtalar

  • ⭐ Statsionar nuqtada: dydx=0\dfrac{dy}{dx} = 0
  • ⭐ Ikkinchi hosila orqali tekshirish:
ShartNuqta turi
d2ydx2<0\dfrac{d^2y}{dx^2} < 0maksimum
d2ydx2>0\dfrac{d^2y}{dx^2} > 0minimum
d2ydx2=0\dfrac{d^2y}{dx^2} = 0aniq emas — dydx\dfrac{dy}{dx} ishorasini nuqtaning ikki tomonida tekshiring
  • Gradient ishorasi orqali tekshirish:
ChapdaNuqtadaO'ngdaTuri
++00−-maksimum
−-00++minimum
++00++egilish nuqtasi (inflection)
−-00−-egilish nuqtasi (inflection)

Barcha xx uchun o'suvchi ekanini isbotlash

  • Barcha xx uchun o'suvchi ⇔ barcha xx uchun f′(x)≥0f'(x) \ge 0 (yoki >0> 0)
  • f′(x)f'(x) kvadrat uchhad bo'lsa: bosh koeffitsient musbat va D<0D < 0, yoki to'la kvadratga keltirib f′(x)=a(x−p)2+qf'(x) = a(x - p)^2 + q, q>0q > 0 ekanini ko'rsatasiz.

🟢 Botiqlik va egilish nuqtasi

ShartGrafik shakli
f′′(x)>0f''(x) > 0pastga botiq (∪, "kosacha")
f′′(x)<0f''(x) < 0yuqoriga botiq (∩, "gumbaz")
f′′f'' ishorasini o'zgartiradiegilish nuqtasi

🔵 Kub funksiya y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d

f′(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c, uning diskriminanti 4(b2−3ac)4(b^2 - 3ac):

ShartStatsionar nuqtalar
b2−3ac>0b^2 - 3ac > 0ikkita (maksimum va minimum)
b2−3ac=0b^2 - 3ac = 0bitta (statsionar egilish nuqtasi)
b2−3ac<0b^2 - 3ac < 0yo'q (funksiya monoton)
  • Egilish nuqtasi har doim: x=−b3ax = -\dfrac{b}{3a}