The cubic function defined as fx ax³ bx² cx d a=0

For the cubic function fx ax3 bx2 cx da 0, the definition is given: Let f x be the function y fx

∵f(x)=ax3 bx2 cx d(a≠0),

∴f′（x）=3ax2 2bx c,f''(x)=6ax 2b,

∵f″（x）=6a*(-

b

3a ) 2b=0,

∴Any cubic function is about the point (-

b

3a ,f(-

b

3a )) Symmetry, that is, ① is correct;

∵Any cubic function has a symmetry center, and the "inflection point" is the symmetry center,

∴There is a cubic function f′(x)=0 with a real solution x0, and the point (x0, f(x0)) is the symmetry center of y=f(x), that is, ② is correct;

Any cubic function has one and only one center of symmetry, so ③ is incorrect;

∵g′（x）=x2-x,g″（x）=2x-1,

Let g″(x)=0, we can get x=

1

2, ∴g(

1

2 )=-

1

2 ,

∴g(x)=

1

3 x3-

1

2 x2-

5

The center of symmetry of

12 is (

1

2 ,-

1

2 ),

∴g(x) g(1-x)=-1,

∴g(

1

2013 ) g(

2

2013 ) … g(

2012

2013 )=-1*1006=-1006, so ④ is correct.

So the answer is: ①②④.

For the cubic function fx ax 3 bx 2 cx da 0, the definition is given: Let f x be the function fx

①From f(x)=2x 3 -3x 2 -24x 12, we get f ′ =6x 2 -6x-24,f ′′ (x)=12x-6.

From f ′′ (x)=12x-6=0, we get x=

1

2 . f(

1

2 )=2*(

1

2 ) 3 -3*(

1

2 ) 2 -24*

1

2 12=-

1

2 .

So the symmetry center coordinate of the function f(x)=2x 3 -3x 2 -24x 12 is (

1

2 ,-

1

2 ) .

So the answer is (

1

2 ,-

1

2 ) .

②Because the symmetry center coordinate of the function f(x)=2x 3 -3x 2 -24x 12 is (

1

2 ,-

1

2 ) .

So f(

1

2013 ) f(

2012

2013 )=f(

2

2013 ) f(

2011

2013 )=…=2f(

1

2 )=2*(-

1

2 ) =-1.

by f(

2013

2013 )=f(1)=-13 .

So f(

1

2013 ) f(

2

2013 ) f(

3

2013 ) … f(

2012

2013 ) f(

2013

2013 ) =-1006-13=-1019.

So the answer is -1019.

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