Question

помогите решить значение функции

Answer 1

$f(x ) = \frac{1}{3} {x}^{3} - \frac{3}{2} {x}^{2} + 5$
f'(x) =
$\frac{1}{3} \times 3 {x}^{2} - \frac{3}{2} \times 2x + 0 = {x}^{2} - 3x$
${x}^{2} - 3x = 0 \\ x(x - 3) = 0 \\ x = 0 \\ x - 3 = 0 \\ x = 3$
Xmax = 0
f(0) = 5 --- наибольшее значение

Xmin = 3
f(3) = 1/3 * 3^3 - 3/2 * 3^2 + 5 = 9 - 27/2 + 5 = 14 - 27/2 = 1/2 = 0,5 -- наименьшее значение

f(-1) = 1/3 * (-1)^3 - 3/2 * (-1)^2 + 5 = -1/3 - 3/2 + 5 = -2/6 - 9/6 + 30/6 = 19/6

f(4) = 1/3 * 4^3 - 3/2 * 4^2 + 5 = 64/3 - 24 + 5 = 64/3 - 19 = 7/3

Ответ: 0,5; 5.