y 3x 2 2x 3。Derivative, extremum increase and decrease table

1.Derivative: y'=6x+6x^2

2.Extremum: Let y'=0, x=0 or -1 and increase or decrease at -1: from 2, the increase interval (negative infinity, -1) and the upper (0, positive infinity) subtract interval [-1, 0].

y = 3x^2 + 2x^3

y' = 6x + 6x^2

y' = 0 ==> 6x + 6x^2 = 0 ==> x = - 1，x = 0

x < 1, x = - 1, - 1 < x < 0, x = 0, x > 0

y'Valid values: +0 - 0 +

In the case of y: Increasing Maximum, Decreasing, Minimum, Increasing.

Maximum: y(-1) = 1

Minimum: y(0) = 0

The derivation is: y=6x+6x 2 has extrema on x=0 and x=-1, when x=-1, there is a maximum y=1, and when x=0, there is a minimum y=0; The function increases from negative infinity to -1, decreases from -1 to 0, and increases from 0 to positive infinity.

Derivative: y'=6x+6x^2

Lingy'=0, then x=0 or x=-1

When x<-1, y'>0, the original function increases monotonically;

When -1 x 0, y'<0, the original function is monotonically decreasing;

When x>0, y'>0, the original function increases monotonically;

So, when x=-1, y obtains a maximum, y=1;

When x=0, y obtains a minimum, y 0;

y'=2(x-3)(x-2)+(x-3) =(x-3)(3x-7)=0, we get x=3, 7 3

y"=(3x-7)+3(x-3)=6x-16 When x=3, y"=18-16>0, so x=3 is the minimum point;

When x=7 3, y"=14-16<0, so x=7 3 is the maximum point.

y=(x-2)(3-x)/x²

i.e. (y+1)x-5x+6=0.

(-5) -4 6 (y+1) 0, solution, y 1 24

When x=12 5, the maximum value y|max=1/24。