Question

Знайти найбільші і найменші значення функції
$f(x) = (x^{2} - 1) \times (x + 1)$
[-2;0]​

Answer 1

$f(x) = ( {x}^{2} - 1)(x + 1) = {x}^{3} + {x}^{2} - x - 1 \\ f'(x) = 3 {x}^{3 - 1} + 2 {x}^{2 - 1} - x {}^{1 - 1} = 3 {x}^{2} + 2x - 1 \\ 3 {x}^{2} + 2x - 1 = 0 \\ a =3\\ b =2\\ c = - 1\\ D =  {b}^{2}  - 4ac = {2}^{2} - 4 \times 3 \times ( - 1) = 4 + 12 = 16 \\ x_{1} = \frac{ - 2 - 4}{2 \times 3} = - \frac{6}{6} = - 1 \\ x_{2} = \frac{ - 2 + 4}{2 \times 3} = \frac{2}{6} = \frac{1}{3} \\ + + + + [ - 1] - - - - [ \frac{1}{3} ] + + + + \\ x_{max} = - 1 \\ x_{min} = \frac{1}{3} \\ \\ y( - 2) = ( - 2) {}^{3} + ( - 2) {}^{2} - ( - 2) - 1 = \\ = - 8 + 4 + 2 - 1 = - 3 \\ \\ y( - 1) = ( - 1) {}^{3} + ( - 1) {}^{2} - ( - 1) - 1 = - 1 + 1 + 1 - 1 = 0 \\ \\ y(0) = {0}^{3} + 0 {}^{2} - 0 - 1 = - 1$

Ответ: у max = 0 ; y min = - 3