y x 3 2 x 5 1

Ответ от Владимир Брест[гуру]
y=(x-3)^2(x+1)
1 ) y´=[ (x-3)^2(x+1) ] ´ = 2* (x-3)*(x+1)*(x-3)´ + (x-3)^2 *(x+1)´
y´ = 2* (x-3)*(x+1) + (x-3)^2
2 ) y´´ = [ 2* (x-3)*(x+1) + (x-3)^2 ] ´ = 2*[ (x-3)*(x+1) ] ´ + 2* (x-3) * (x-3)´
y´´ = 2 * [ (x-3)´ * (x+1) + (x-3)*(x+1)´ ] +2 * (x-3)
y´´ = 2 * [ (x+1) + (x-3)] +2 * (x-3)
y´´ = 2 * [ (x+1) + (x-3)] +2 * (x-3)
y´´ = 2 * ( 3 * x - 5 )
3 ) y ´= 0 ;
2* (x-3)*(x+1) + (x-3)^2 = 0
(x-3)*(3*x -1) = 0
X1 = 3 ; X2 = 1 /3
4 ) y´´( 3 ) = 2 *( 3 * 3 -5 ) = 8 ; y´´( 3 ) = 8 > 0 >>> в точке x = 3 наблюдается min ; f ( 3 ) = 0
5 ) y´´( -1,5 ) = 2 *[ 3 * (1/3) -5 ] = -9< 0 >>> в точке x = 1 /3 наблюдается max ; f (1 /3) = 9,48