§ 41: 41.9 (1)

41.9 Знайдіть найбільше і найменше значення функції f на вка–

заному відрізку:

1) f(x) = 2sin2x + cos4x;

f’(x) = 4cos2x – 4sin4x = 4(cos2x – 2sin2xcos2x) – 4cos2x(1 – 2sin2x).

f(x) = 0,4cos2x(1 – 2sin2x) = 0;

cosx = 0 2x = π2 + πk, x = π2 + πk2,

sin2x = 12; 2x = (–1)^k π6 + πk, k ∈ Z; x = (–1)^k π12 + πk2, k ∈ Z;

Точки такого виду є критичними, до відрізка [0; π3] входять точки

х = π4, х = π12.

f(π4) = 2sin2π4 + cos4π4 = 2sinπ2 + cosπ = 2 – 1 = 1;

f(0) = 2sin0 + cos0 = 1;

f(π3) = 2sin2π3 + cos4π3 = 232 – 32 = 32;

f(π12) = 2sin2π12 + cos4π12 = 2sinπ6 + cosπ3 = 2 • 12 + 12 = 32;

maxf(x) = f(π12) = 32; minf(x) = f(π4) = 1.

_[0; π3_{] [0;} π3_]

2) f(x) = 3sin2x + cos2x – 5; f’(x) = 23cosx – 2sinx.

f’(x) = 0; 23cos2x – 2sin2x = 0; 3cos2x = sin2x; tg2x = 3;

2x = π3 + πk, k ∈ Z; x = π6 + πk2, k ∈ Z.

f(π6) = 3sin2π6 + cos2π6 – 5 = 3sinπ3 + cosπ3 – 5 = 3 • 32 + 12 – 5 = –3.

f(0) = 3sin0 + cos0 – 5 = 1 – 5 = –4;

f(π3) = 3sin2π3 + cos2π3 – 5 = 3 • 32 – 12 – 5 = 32 – 12 – 5 = –3;

maxf(x) = f(π6) = –3; minf(x) = f(0) = –4.

_[0; π3_{] [0;} π3_]