y is defined by the cubic function fx=ax^3+bx^2+cx+d

For the cubic function fx ax 3 bx 2 cx da 0 definition: Let f x be the derivative y of the function y fx

(1) According to the meaning of the question, we get: f′(x)=3x 2 -12x 5, ∴f′′(x)=6x-12=0, we get x=2

So the inflection point coordinate is (2,-2)

(2) Assume (x 1 , y 1 ) and (x, y) are symmetrical about the center of (2,-2), and (x 1 , y 1 ) is at f(x), so there is

x 1 =4-x

y 1 =-4-y ,

From y 1 =x 1 3 -6x 1 2 5x 1 4, we get -4-y=(4-x) 3 -6(4-x) 2 5(x-4) 4

Simplified: y=x 3 -6x 2 5x 4

So (x, y) is also on f(x), so f(x) is symmetric about the point (2,-2).

The "inflection point" of the cubic function f(x)=ax 3 bx 2 cx d(a≠0) is (-

b

3a ,f(-

b

3a )), which is the center of symmetry of function f(x)

(Or: any cubic function has an inflection point; any cubic function has a center of symmetry; any cubic function can be an odd function after translation).

(3),G(x)=a(x-1) 3 b(x-1) 2 3(a≠0), or write a specific function, such as G(x)=x 3 -3x 2 3x 2, or G(x)=x 3 -3x 2 5x

For the cubic function fx ax3 bx2 cx da 0 definition: let f x be the derivative of function y fx

(1)f′(x)=3x2-6x 2…(1 point) f″(x)=6x-6 Let f″(x)=6x-6=0 and get x=1…(2 points) )f(1)=13-3 2-2=-2∴Inflection point A(1,-2)…(3 points)

(2) Suppose P(x0,y0) is any point on the image of y=f(x), then y0=x03-3x02 2x0-2, because P(x0,y0) is about A(1,-2 ) is P'(2-x0,-4-y0),

Substituting P' into y=f(x), we get the left side =-4-y0=-x03 3x02-2x0-2

Right side=(2-x0)3-3(2-x0)2 2(2-x0)-2=-x03 3x02-2x0-2∴Right side=Right side∴P′(2-x0,-4- y0) On the graph of y=f(x), ∴y=f(x) is symmetric about A... (7 points)

Conclusion: ①The inflection point of any cubic function is its center of symmetry

②Any cubic function has an "inflection point"

③Any cubic function has a "center of symmetry" (write one of them)...(9 points)

(3) Suppose G(x)=ax3 bx2 d, then G(0)=d=1...(10 points) ∴G(x)=ax3 bx2 1,G'(x)=3ax2 2bx,G ''(x)=6ax 2bG''(0)=2b=0,b=0, ∴G(x)=ax3 1=0...(11 points)

Fa1:

G(x1) G(x2)

2 ?G(

x1 x2

2 )=

a

2

x 3

1

a

2

x 3

2

?a(

x1 x2

2 )3=a[

1

2

x 3

1

1

2

x 3

2

?(

x1 x2

2 )3]=

a

2 [

x 3

1

x 3

2

?

x 3

1

x 3

2

3

x 2

1

x2 3x1

x 2

2

4 ]=

a

8 (3

x 3

1

3

x 3

2

?3

x 2

1

x2?3x1

x 2

2

)=

a

8 [3

x 2

1

(x1?x2)?3

x 2

2

(x1?x2)]=

3a

8 (x1?x2)2(x1 x2)…(13 points)

When a>0,

G(x1) G(x2)

2 >G(

x1 x2

2 )

When aG(x1) G(x2)

2 x1 x2

2)…(14 points)

Method 2: G′′(x)=3ax, when a>0, and x>0, G′′(x)>0, ∴G(x) is a concave function at (0, ∞) ,∴

G(x1) G(x2)

2 >G(

x1 x2

2 )…(13 points)

When aG(x1) G(x2)

2 x1 x2

2)…(14 points)

For the cubic function fx ax3 bx2 cx da 0 definition: let f x be the derivative function of function y fx

(1)∵f'(x)=3x2-6x 2,

∴f''(x)=6x-6,

Let f''(x)=6x-6=0,

Get x=1,f(1)=-2

So the coordinates of "inflection point" A are (1,-2)

(2) Suppose P(x0,y0) is any point on the image of y=f(x), then y0=x03?3x02 2x0?2

∴P(x0,y0) is symmetric about (1,-2) point P'(2-x0,-4-y0),

Substituting P'(2-x0,-4-y0) into y=f(x), what is the left side =? 4?y0=?x03 3x02?2x0?2

Right side=(2?x0)3?3(2?x0)2 2(2?x0)?2=?x03 3x02?2x0?2

∴Left side=right side,

∴P'(2-x0,-4-y0) on the y=f(x) image,

The image of ∴f(x) is symmetrical about the "inflection point" A.

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