Розділ 4. Функції » 1

Дано функції. Порівняйте: f(x) = –3x2 + 5x + 1; g(x) = 2 + x2 + 3x; 1) f(1) < g(1). f(1) = –3 • 12 + 5 • 1 + 1 = –3 + 5 + 1 = 3; g(1) = 2 + 12 + 3 • 1 = 2 + 1 + 3 = 6. 2) f(2) < g(–2); f(2) = –3 • 22 + 5 • 2 + 1 = –12 + 10 + 1 = –1; g(–2) = 2 + 2 • (–2)2 + 3 • (–2) = 2 + 8 – 6 = 4. 3) f(–1) < g(0); f(–1) = –3 • (–1)2 + 5 • (–1) + 1 = –3 – 5 + 1 = –7; g(0) = 2 + 02 + 3 • 0 = 2 + 0 + 0 = 2.