Question

Найдите наибольшее и наименьшее значения функции на отрезке:
f(x)=x^3-3x^2-9x+23; [-4;4]

Answer 1

Ответ:

f(x)=x^3-3x^2-9x+23;

f'(x)=(x^3-3x^2-9x+23)' = 3x²-6x-9 {:3} = x²-2x-3.

Находим корни уравнения

 x²-2x-3=0.

По т. Виета

x1+x2=2;

x1*x2=-3;

x1=-1;

x2=3.

Точки экстремума  x1=-1;  x2=3.

MAX => f(-1) = (-1)³-3(-1)²-9(-1)+23 = 28.

MIN =>  f(3) =  (3)³-3(3)²-9(3)+23 = -4.

Answer 2

$f(x) = x {}^{3} - 3 {x}^{2} - 9x + 23 \\ f'(x) = 3 {x}^{2} - 6x - 9 = 3( {x}^{2} - 2x - 3) \\ {x}^{2} - 2x - 3 = 0 \\ a =1 \\ b = - 2 \\ c = - 3 \\ D = {b}^{2} - 4ac = ( - 2) {}^{2} - 4 \times 1 \times ( - 3) = 4 + 12 = 16 \\ x_{1} = \frac{2 - 4}{2} = - \frac{2}{2} = - 1 \\ x_{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3 \\ + + +[ - 1] - - - [3] + + + \\ \\ y( - 4) = ( - 4) {}^{3} - 3 \times ( - 4) {}^{2} - 9 \times ( - 4) + 23 = \\ = - 64 - 3 \times 16 + 36 + 23 = - 5 - 48 = - 53 \\ \\ y( - 1) = ( - 1) {}^{3} - 3 \times ( - 1) {}^{2} - 9 \times ( -1 ) + 23 = \\ = - 1 - 3 + 9 + 23 = 28\\ \\ y(3) = {3}^{3} - 3 \times {3}^{2} - 9 \times 3 + 23 = 27 - 27 - 27 + 23 = \\ = - 4 \\ \\ y(4) = {4}^{3} - 3 \times {4}^{2} - 9 \times 4 + 23 = 64 - 3 \times 16 - 36 + 23 = \\ = 51 - 48 = 3$

Ответ: у min = - 53 ; y max = 28