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Aralash mashqlar: hosila va ko‘p bosqichli masalalar (1.13)
maktab darsligi: matematika 11-sinf; 22–23-hafta mavzularini qayta ko‘ring
Maqsad: Hosila yordamida urinma, kesmada eng katta va eng kichik qiymat va amaliy ekstremal masalalarni yechish.
Qoida va g‘oya
Eslatma: (x^n)' = nx^(n-1), (uv)' = u'v + uv', (u/v)' = (u'v - uv')/v^2, murakkab funksiyada tashqi hosila ichki hosilaga ko‘paytiriladi. Urinma: y = f(x0) + f'(x0)(x - x0); parallel urinmada f'(x0) = k; normalning og‘ishi -1/f'(x0). Ekstremum: f' ishorasi + dan - ga o‘tsa maksimum. Kesmada eng katta va eng kichik qiymat: kritik nuqtalar (kesma ichidagilari) va chetlarda qiymatlarni solishtiring. Ekstremal masalada o‘zgaruvchilarni shart bo‘yicha bittaga keltiring, aniqlanish oralig‘ini yozing. Tezlik v = s', tezlanish a = v'.
Namunaviy masalalar
f(x) = x^3 - 3x^2 + 2 grafigiga x0 = 1 nuqtada o‘tkazilgan urinma y o‘qini qaysi nuqtada kesadi?
Yechim: f(1) = 0, f' = 3x^2 - 6x, f'(1) = -3. Urinma: y = -3(x - 1) = -3x + 3. x = 0 da y = 3.
Javob: (0; 3)
Yechim: f(1) = 0, f' = 3x^2 - 6x, f'(1) = -3. Urinma: y = -3(x - 1) = -3x + 3. x = 0 da y = 3.
Javob: (0; 3)
f(x) = x^3/3 - 2x^2 + 3x funksiyaning [0; 4] kesmadagi eng katta va eng kichik qiymatini toping.
Yechim: f' = x^2 - 4x + 3 = (x - 1)(x - 3). f(0) = 0, f(1) = 4/3, f(3) = 0, f(4) = 64/3 - 32 + 12 = 4/3. Eng katta 4/3, eng kichik 0.
Javob: max 4/3, min 0
Yechim: f' = x^2 - 4x + 3 = (x - 1)(x - 3). f(0) = 0, f(1) = 4/3, f(3) = 0, f(4) = 64/3 - 32 + 12 = 4/3. Eng katta 4/3, eng kichik 0.
Javob: max 4/3, min 0
Kvadrat asosli yopiq qutining hajmi 64 dm^3. Uning to‘la sirti eng kichik bo‘lganda sirt yuzini toping.
Yechim: Asos tomoni x, balandlik h = 64/x^2. S = 2x^2 + 4xh = 2x^2 + 256/x. S' = 4x - 256/x^2 = 0, x^3 = 64, x = 4, h = 4. S = 32 + 64 = 96.
Javob: 96 dm^2
Yechim: Asos tomoni x, balandlik h = 64/x^2. S = 2x^2 + 4xh = 2x^2 + 256/x. S' = 4x - 256/x^2 = 0, x^3 = 64, x = 4, h = 4. S = 32 + 64 = 96.
Javob: 96 dm^2
Tipik xatolar
- Urinma tenglamasida f'(x0) o‘rniga f'(x) ni qoldirib qo‘yish.
- Kesma chetlarini hisoblashni unutish: ekstremum nuqtasidagi qiymat eng katta bo‘lmasligi mumkin.
- Ekstremal masalada ikkinchi o‘zgaruvchini yo‘qotmay hosila olish.
Mashq
1
y = x^2 - 2x + 3 grafigiga y = 4x - 1 ga parallel urinma tenglamasini yozing.
y = 4x - 6
2
s(t) = t^3 - 6t^2 + 9t. Tezlik nolga teng bo‘ladigan ikki moment orasida jism yo‘lining koordinata o‘zgarishi moduli nechaga teng?
4
3
f(x) = 2x^3 - 3x^2 - 36x + 10 ning minimum nuqtasini toping.
3
4
sqrt(36,12) ning taqribiy qiymatini toping.
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