§ 41: 41.1 (1-2)

41.1 Знайдіть найбільше і найменше значення функції f на вказаному відрізку:

1) f(x) = Зх² – x³; f’(x) = 3 • 2x – Зх² = 6x – Зх² = Зх(2 – x).

f’(x) = 0; 3x(2 – x) = 0; x = 0 або x = 2.

f(0) = 3 • 0 – 0 = 0; f(2) = 3 • 2² – 2³ = 12 – 8 = 4;

f(–1) = 3 • (–1)² – (–1)³ = 3 + 1 = 4;

f(3) = 3 • З² – З³ = 27 – 27 = 0;

maxf(x) = f(2) = f(–1) = 4; minf(x) = f(0) = f(3) = 0.

^{[–1; 3] [–1;3]}

2) f(x) = x⁴ – 2x² + 5; f’(x) = 4x³ – 4x = 4x(x² – 1) = 4x(x – 1)(x + 1).

4x(x – 1)(x + 1) = 0 при x₁ = 0, x₂ = 1, x₃ = –1. З цих критичних

точок відрізку [0; 2] належать дві: 0 i 1.

f(0) = 0⁴ – 2 • 0 + 5 = 5;

f(1) = 1⁴ – 2 • 1 + 5 = 1 – 2 + 5 = 4;

f(2) = 2³ – 2 • 2² + 5 = 16 – 8 + 5 = 13; maxf(x) = f(2) = 13;

min f(x) = f(1) = 4. ^{[0; 2]}

^{[0; 2]}

3) f(x) = 2x³ + 9x² – 60х – 7; f’(х) = 6х² + 18x – 60.

6х² + 18х – 60 = 0; x² + Зх – 10 = 0; х₁ = 2; x₂ = –5;

Відрізку [–1; 3] належить лише критична точка х = 2.

f(–1) = 2 • (–1)³ + 9 • (–1)² – 60 • (–1) – 7 = –2 + 9 + 60 – 7 = 60;

f(3) = 2 • 3³ + 9 • З² – 60 • 3 – 7 = 54 + 81 – 180 – 7 = –52;

f(2) = 2 • 2³ + 9 • 2² – 60 • 2 – 7 = 2 • 8 + 9 • 4 – 120 – 7 = 16 + 36 – 127 = –75;

max f(x) = f(–1) = 60; min f(x) = f(2) = –75.

^{[–1; 3] [–1; 3]}

4) f(x) = x2+ 8x-1; f’(x) = x2+ 8'x-1- (x2- 8)(x-1)'(x-1)2 = 2xx-1- (x2+ 8)(x-1)2 = x2- 2x- 8(x-1)1.

f’(x) = 0 при х = 4 і х = –2, f’(x) не існує при х = 1. До відрізка [–3; 0] належить тільки критична точка х = –2.

f(–2) = (-2)2+ 8-2-1 = 4 +8-3 = –4; f(–3) = (-3)2+ 8-3-1 = 17-4 = –174;

f(0) = 0+80-1 = –8;

max f(x) = f(–2) = –4; min f(x) = f(0) = –8.

^{[–3; 0] [–3; 0]}