Question

Найти наибольшее и наименьшее значение функции f(x)=х^3- 3х^2 -45 х на отрезке [-2;6]

Answer 1

$f(x) = {x}^{3} - 3 {x}^{2} - 45x \\ f'(x) = 3 {x}^{3 - } - 3 \times 2 {x}^{2 - 1} - 45 {x}^{1 - 1} = 3 {x}^{2} - 6x - 45 \\ 3 {x}^{2} - 6x - 45 = 0 \\ {x}^{2} - 2x - 15 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\{x}^{2}+ bx + c = 0 \\ x_{1}+x_{2} =- b\\ x_{1}x_{2}=c\\ \\ x_{1}+x_{2}=2\\ x_{1}x_{2} =-15\\ x_{1} =-3\\ x_{2} =5\\ \\ + + + + [ - 3] - - - - [5] + + + + \\ x_{max} = - 3 \\ x_{min} = 5 \\ \\ y( - 2) =( - 2) {}^{3} - 3 \times ( - 2) {}^{2} - 45 \times ( - 2) = \\ = - 8 - 3 \times 4 + 90 = 82 - 12 = 70 \\ \\ y(5) = {5}^{3} - 3 \times {5}^{2} - 45 \times 5 = 125 - 3 \times 25 - 225 = \\ = - 100 - 75 = - 175 \\ \\ y (6) = {6}^{3} - 3 \times {6}^{2} - 45 \times 6 = 216 - 3 \times 36 - 270 = \\ = - 54 - 108 = - 162$

Ответ: y max = 70 ; y min = - 175