Let the function f x x 3 bx 2 cx x R , and we know that g x f x f x is an odd function

f'(x)=3x^2+2bx+c

So g(x)=x 3+(b-3)x 2+(c-2b)x-cg(-x)=-x 3+(b-3)x 2-(c-2b)x-c is an odd function. g(-x)=-g(x)

x^3+(b-3)x^2-(c-2b)x-c=-x^3-(b-3)x^2-(c-2b)x+c

2(b-3)x^2-2c=0

b-3=0,c=0

b=3,c=0

So g(x)=x3-6x=m

For the equation: x 3-6x-m=0

If there are at least two different real roots, then the derivative of the function is required to be: 3x 2-6=0, so the minimum and maximum values are obtained at x=- 2 2 and x= 2 2, in order to ensure that the function image and the x-axis have at least two intersections, then when x=- 2 2, the function value must be greater than or equal to 0, that is:

Bring in x=- 2 2 to get:

11√2/4-m>0

Solution: m<11 2 4

f(x)=x^3+bx^2+cx

f‘(x)=3x^2+2bx+c

g(x)=x 3+(b+3)x 2+(2b+c)x+c odd function So b+3=0 c=0

b=-3g(x)=x 3-6x=x(x 2-6)=0 x=o or 6

g(x)'=3x 2-6=0 x = 2, so the function image is wavy. At x 6 - 6 x 0 y is positive at x 6 x 0 y is negative.

where x= 2 is the tangent point and x= 2 is y= 4 2 as seen in the image.

m∈〔-4√2,4√2〕

1 All (1) solution: Since g(x) is an odd function, there is g(0)=0, i.e., f(0)+f'(0)=0, the solution is b=0.

And because there is an equation g(x)=-g(-x), the solution is (3a+1)x 2=0. Let 3a+1=0, and the solution is a=-1 3.

So f(x)=-x 3 3+2x.

2) Solution: From (1), g(x)=-x 3+2x. Finding a derivative of it yields g'(x)=x^2+2〉0。

So the function g(x) is an increasing function in the interval. So the maximum value of g(x) over the interval [1,2] is g(2)=6 and the minimum value is g(1)=3

f'(x)=3ax^2+2x+b

g(x)=f(x)+f'(x)=ax^3+(3a+1)x^2+(b+2)x+b

g(x)=f(x)+f'(x) is an odd function.

g(x)=g(-x)

So 3a+1=0 a=-1 3

The expression b = 0f(x) -1 3x 3+x 2g(x) = -1 3x 3+2x

g(x)'=-x^2+2

So increase the interval [-root number 2, root number 2].

Subtract interval (negative infinity, - root number 2] [root number 2, positive infinity) is in the interval [1,2].

Increase first and then decrease. Maximum g (root number 2) = 3 2

The minimum value g(2) = -2

g(x) = f(x)-f'(x)

x +bx +cx-(3x +2bx+c) = x +(b-3)x +(c-2b)x-c g(x) is an odd function.

g(-x)=(-x)³+b-3)(-x)²+c-2b)(-x)-c

x + (b-3) x - (c-2b) x-c = -g(x) = -x - (b-3) x - (c-2b) x + c contrast coefficient, obtain: b-3 = 3-b, -c=c

b=3，c=0

b=-3,c=0

g(x)-f(x)-f'(x) is an odd function.

The description is that the odd function plus the odd function is equal to the odd function.

Even the chain of even talk is even.

The sum of odd-even functions is non-odd and non-even).

Root order f(x) g(x)-f(x)-f'(x), f (x) f(x) 0 can be obtained from the odd function, slowly resolved, according to the principle of one-to-one correspondence, it can contain chain sun c to call 0, 2b 6 0, b 3

b=3,c=0

g(x)=x3+(b-3)x2+(c-2b)x-c, because the sideslip is an odd function must pass (0,0) points, so the substituting c=0 is g(x)=x3+(b-3)x2-2bx

And because of the odd letter noisy oak number, there is luck to touch wax g(-x)=-g(x), that is, (-x)3+(b-3)(-x)2-2b(-x)=-x3-(b-3)(-x)2+2bx

The solution is b=3

∵f′(x)=3x²+2bx+c

g(x)=f(x)-f'(x)

x³+bx²+cx－3x²－2bx－c

x³＋﹙b－3﹚x²＋﹙c－2b﹚x－c∵g(x)=f(x)-f'(x) is an odd function.

b－3＝0 ,c＝o

b＝3, c＝0

f (x)=3x +6x 3x x 2 is obtained by 3x x 2 0.

x 0 or x 2

0 and 2 are extreme points.

The maximum is f(0)=0 and the minimum is f2 4

Find the derivative of f(x) and add it to f(x) to get g(x), and then use g(-x) = -g(x); g(0)=0, get the values of b, c, and then the extreme values are easy to find.

It should be funny fx=x 3+bx 2+cx, and the mountain sail is absolutely unique.

gx=f(x)-f'(x)

x^3+bx^2+cx-3x^2-2bx-cgx=f(x)-f'(x) is an odd function.

Then the sedan g(-x)=-x 3+bx 2-cx-3x 2+2bx-c=-g(x)=-x 3-bx 2-cx+3x 2+2bx+c

So b=3, c=0

Because f(x) is an odd function, there is a rock grip f(-x)=-f(x) i.e. -x) 3+b(-x) 2+c(-x)=-x 3-bx 2-cx, so bx 2=-bx 2, and coarse spike thus b=0

The same goes for g(x) = g(-x).

Substituting yields c = 2

The skin foci were b=0, c=2

Thank you

g(x)=f(x)-2=x 3+ax 2+3x+b-2 is an odd function, and if the domain is r, then there is g(0)=b-2=0

Therefore, b = 2g (-x) = -x 3 + ax 2-3x

g(x)=-x^3-ax^2-3x

g(-x)=-g(x)

Therefore ax 2=-ax 2, we get a=0

In summary, a=0, b=2