§ 41: 41.4 (4)

41.4 Знайдіть найбільше і найменше значення функції f на вка–

заному відрізку:

1) f(x) = 9+8x- x2;

f’(x) = 129+8x- x2 • (9 + 8 – x²)’ = 8-2x29+8x- x2 = 4-x9+8x- x2;

f’(x) = 0 при x = 4. –x² + 8x + 0 = 9; x² – 8x – 9 або х = –1.

До проміжку [0; 7] входить критична точка х = 4.

f(0) = 9+8 •0- 02 = 9 = 3; f(7) = 9+8 •7- 72 = 9+56-49 = 16 = 4;

f(4) = 9+8 •4- 42 = 9+32-16 = 25 = 5;

maxf(x) = f(4) = 5; minf(x) = f(0) = 3.

^{[0; 7] [0; 7]}

2) f(x) = 4xx2+ 1;

f’(x) = 4x2+ 1- 4x • 2x(x2+ 1)2 = 4x2+ 4-8x2(x2+ 1)2 = 4-4x2(x+ 1)2 = 41-x(1+x)(x+ 1)2;

f’(x) = 0 при х = 1 або х –1.

f(1) = 4 • 112+ 1 = 42 = 2; f(–1) = -4(-1)2+ 1 = –42 = –2; f(–2) = 4 • (-2)(-2)2+ 1 = –85;

f(4) = 4 • 442+ 1 = 1617; maxf(x) = f(1) = 2; minf(x) = f(–1) = –2.

^{[–2; 4] [–2; 4]}

3) f(х) = (х – 1)²(х + 5)²;

f’(x) = 2(х – 1)(х + 5)² + (х – 1)² • 2(х + 5) = 2(х – 1)(х + 5)(х + 5 + х – 1) = 2(х – 1)(х + 5)(2х + 4) = 4(х – 1)(х + 5)(х + 2).

f’(x) = 0 при х = 1, х = –5, х = –2 – це критичні точки.

Проміжку [–3; 2] належать тільки дві із них: x = 1 i x = –2.

f(1) = (1 – 1)² • (1 + 5)² = 0;

f(–2) = (–2 – 1)² • (–2 + 5)² = 9 • 9 = 81;

f(–3) = (–3 – 1)² • (–3 + 5) = 16 • 4 = 64;

f(2) = (2 – 1)² • (2 + 5)² = 1 • 49 = 49;

maxf(x) = f (–2) = 81; minf(х) = f (1) = 0.

^{[–3; 2] [–3; 2]}

4) f(x) = –x – 9x; f’(x) = –1 + 9x2 = 3-x(3+x)x2.

f’(x) = 0 при х = 3 і х = –3. f’(x) не існує при х = 0.

З трьох критичних точок відрізку [–6; –1] належить одна х = –3.

f(–3) = –(–3) – 9-3 = 3 + 3 = 6; f(–6) = –(–6) – 9-6 = 6 + 32 = 7,5;

f(–1) = –(–1) – 91 = 1 + 9 = 10; maxf(x) = f(–1) = 10;

minf(x) = f(–3) = 6. ^{[–6; –1]}

^{[–6; –1]}