Відповіді АЛГЕБРА: 24.3

24.3

1) f(x) = 3x – 7; [0; 2];

f найм (0) = –7;

g найб (2) = 6 – 7 = –1;

2) f(x) = x² – 2x + 3; [2; 3];

f’(x) = 2x – 2 = 0;

x = 1 ∈ [2; 3];

f(2) = 4 – 4 + 3 = 3 – найм;

f(3) = 9 – 6 + 3 = 6 – найб;

3) f(x) = –x2 + 4x + 5 [1; 4];

f’(x) = –2x + 4 = 0;

x = 2;

f(1) = –1 + 4 + 5 = 8 найб;

f(2) = –4 + 8 + 5 = 9 найм;

f(4) = –16 + 16 + 5 = 5;

4) f(x) = x – 13x³ [–2; 0];

f’(x) = 1 – x²

1 – x² = 0;

x = 1 x = –1;

f(–2) = –2 + 13 • 8 = 83 – 2 = 23 – найб;

f(–1) = –1 + 13 = –23 – найм (0) = 0;