Найти экстрэмум x^3-5x^2+3x-5
Y ' = 3x^2 - 10x + 3 y ' = 0 3x^2 - 10x + 3 = 0 D = 100 - 36 = 64 x1 =( 10 + 8)/6 = 3 x2 = ( 10 - 8)/6 = 1/3 + - + ------- 1/3 -----------3 --------> x f min = y (3) = - 14 f max = y(1/3) = - 122/27