If the functions f x ax 3 bx 2 2x c get the extremum when x 2 and x 1 are obtained. 10

1) f(x)=ax 3+bx 2-2x+c, then f'(x)=3ax^2+2bx-2

and f'The two roots of (x)=0 are x=-2 and x=1

Substituting the solution yields a=1 3, b=1 2

2) By the nature of the cubic function, only the maximum value is greater than 0 and the minimum value is less than 0, i.e., f(-2)>0 and f(1)<0

The value range of the solution c is (-10 3, 7 6).

Landlord.. Are you sure your function is not typographed?

It's f(x)=ax 3-bx 2-2x-c.

The first question: f(x)=ax 3+bx 2-2x+c, then f'(x)=3ax^2+2bx-2

and f'The two roots of (x)=0 are x=-2 and x=1

Substituting the solution yields a=1 3, b=1 2

The second question consists of the properties of the cubic function, only the maximum value is greater than 0 and the minimum value is less than 0, i.e., f(-2)>0 and f(1)<0

The value range of the solution c is (-10 3, 7 6).

Your equal sign should be a plus, right?

f(x) derivative, after derivative, when x=1, x=-2, its derivative is 0, you can find a, b

f(x)=ax^3=bx^2-2x=c?If the = to the right of f= is +, then:

Derivative of f with respect to x, f'=3a x 2+2b x-2, which is a quadratic function, and the extreme point of the function f is its derivative f'=0, i.e. f'The solution of =0 is -2 and 1, and a and b can be found according to the knowledge of quadratic equations

1、f'(x)=3x²+2ax+b=0

Both x=1 and -2 have extremums.

So x=1 and -2 are the roots of the equation.

By the Vedic theorem.

2a/3=-(-2+1)=1

b/3=-2*1=-2

a=3/2,b=-6

2、f'(x)=3x²+3x-6

So -21 is an increment function, so x=1 has a minima.

So f(1)=0

1+3/2-6+c=0

c=7/2

f'The solution of (x)=3x 2+2ax+b=0 is 1, and -2 is substituted into:

3+2a+b=0

12-4a+b=0

a=,b=-6

f(x)=x^3+

f(1)=1+

f(-2)=-8+6+12+c=c+10

Therefore, min f(x)=f(1)=, x [-3,2] is c=

x=1 is the root of df dx = 3x 2 +2ax + b = 0, so 3+2a+b = 0 --1, where f(1) = 1+a+b+a2 = 10 --22-1.

A 2 -A -12=0, A = 4 or A=-3 When A = 4, B = -11, Df Dx = 3X 2 +8X -11, when A = -3, B = 3, Df Dx = 3X 2 -6X +3, in this case since the second derivative is also 0 at X = 1, it is not an extreme point.

So a=4, b=-11

Bring it in to get f(2).

Solution: f(x)=x 3+ax 2+bx+a 2 has an extreme value at x=1 10 then f(1)=10=1+a+b+a f'(1)=0=3+2a+b

a=-3, b=3 or a=4, b=-11f(2)=11 or 10

From the expression f(x) we get f'(x) = 3x^2+2ax+b，f''(x)=6x+2a

From the extreme value 10 at x=1 f(x), we can see that f(1)=1+a+b+a 2=10, i.e., a+b+a2=9....

and f'(1)=3+2a+b=0...

Combine and solve a=-3, b=3 or a=4, b=-11

When a=-3, b=3, f''(1) = 0, (1, 10) is not an extreme point, it is not in line with the topic, and it is discarded.

When a=4, b=-11, f''(1) = 14>0, (1, 10) is the minimum point, which is in line with the title.

Substituting a=4, b=-11 into f(x), we can see that f(2)=18

2.Because.

f(x)=ax 3+bx 2-3x=x times 2=2x, so for the values x1,x2 of any two independent variables on the interval [-1,1], there is |f(x1)-f(x2)|≤4

Nothing else will be, I haven't learned it yet!

Thank you for adopting

The first question on the first floor has been solved, and the second question is f'(x)=6x 2-12x-18=6(x+1)(x-2), it can be seen that f(x) is monotonically increasing in (— 1), (-1,2) monotonically decreasing, 2,+ monotonically decreasing, the meaning of the problem is that in the interval [m,m+4] is f(x) is a monotonic function, which may be monotonically increasing or monotonically decreasing, but it must be monotonic, that is, in [this interval, the sign of f'(x) must be constant positive or negative, and cannot be changed, and the length of this interval is 4, However, the monotonically decreasing interval is (-1,2) and the length is 3, so f(x) cannot be monotonically decreasing on [, so it can only be monotonically increased. So m+4 -1, or m2, so the result is m-2, or m2. Let's make it clear ...

1.From the meaning of the title f'(x)=6x2+2ax+b should be 0 at x=-1 and x=2, substitute them for ,--a=-6, b=-18

f(x)=2x 3-6x 2-18x+3, the maximum is: f(-1)=-5, f(2)=-41

2. f'(x)=6x2-12x-18=6(x+1)(x-2), m+4<=-1 or m>=2---m<=-5 or m>=2

The derivative is equal to zero, and the standing point is found, and according to the interval and the value range of a and b, the extreme value and the m value range can be solved.