Відповіді АЛГЕБРА: 24.17

24.17

1) f(x) = x2+ 4x-1x-1; [2; 4];

D(y) x ≠ 1;

f’(x) = 2x+4x-1- (x2+ 4x-1)(x-1)2 = 2x2-2x+4x-4- x2- 4x+1(x-1)2 =

= x2- 2x-3(x-1)2

f’(x) = 0;

x² – 2x – 3 = 0;

x₁ = –1; x₂ = 3;

f(2) = 4+8-12-1= 111= 11 – найб. зн.;

f(3) = 9+12-13-1 = 202 = 10 – найм. зн.;

f(4) = 16+16-14-1 = 313 = 1013 ;

2) f(x) = xx+3 ; [–3; 1];

f’(x0 = x+3 + x • 12x+3 = 2x+3+ x2x+3 = 3x+62x+3 = 0; x = –2;

f(–3) = –30 = 0;

f(–2) = –21 = –2 – найменше;

f(1) = 1 • 4 = 2 – найбільше.