It is known that the monotonic reduction interval of the function f x 3 ax 2 bx 3 is 1, 3, 1 .

1。Do it down. Bring in the endpoint value.

2。Discriminant equations for quadratic equations.

After the derivation, it can be known. x 2+ax+b, that is, at the zero point of this function, 1 3, we know a=1 12

b=-1/12 .I don't know if I miscalculated ......

The second one is a tower, 64-8t positive number two, negative number misunderstanding, 0 one.

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

When x = -1 3 and x = 1 f'(x）=0

a=-1 and b=-1 can be obtained

f(x)=x^3-2x^2-x+3

2)∵f(x)=x^3-2x^2-x+3

x 3-2x 2-x+3=2x 2+8x+t, simplified to x 3-4x, 2-9x+3-t=0

Let the left side be g(x) and find the derivative so that it is 0

0, g'(x) 0 is constant, so g(x) is an incrementing function.

So there is only one real root.

That's what I think, please combine your views

1.Derivative, because the extreme point equation 1 3 and 1 so f'（－1／3），f'(1) is equal to 0, and the solution is a=b=-1

2.After simplification, we get x 3-3x 2-9x+3=t, find the derivative on the left side of the equation, you can get the maximum and minimum value, draw the graph, use the line y=t to connect the image, and the function on the left is monotonically increased by the derivative, so there is only one real root.

f'(x)=x 2ax frontal circle b, then f'(x) satisfies on interval [1,2]: f'( 1)=1 2a b mountain good 0; ②f'(2)=4＋4a－b≤0.So your problem is to find the maximum value of the objective function z=a b over the region, which is a linear programming comma lead problem.

Solve this problem, you can.

f(x)=x 3+ax 2+bx+c is a monotonically decreasing function in the interval [-1,0], then f (x)=3x 2+2ax+b is invariably less than or equal to 0 in the interval [-1,0], and the image of the quadratic function 3x 2+2ax+b is drawn, and we can see that f (-1) 0, f (0) 0, that is, 3-2a+b 0, b 0.......

With a as the horizontal axis and b as the vertical axis, the feasible domain represented by the formula *) is the common part below the right of the line 3-2a+b=0 and below the b=0 (i.e., y-axis), (a 2+b 2) represents the distance from the origin to the feasible domain, the minimum value is the distance from the origin to the straight line -2a+b+3=0, which is 3 5, and the minimum value of a 2+b 2 is 9 5

If a=3, then f(x)=-1 3x 3+x 2+3x+bf'(x)= x^2+2x+3

Order f'(x)=0, the solution is x1=-1, x2=3 in the area between the lead x (-1), f'(x)<0.The function is monotonic and the eggplant is diminishing.

In the interval x [-1,3),f'(x)>0.Functions are monotonically incremented.

In the interval x [3, + f'(x)<0.Functions are monotonically decreasing.

For the elementary derivative function to find the monotonic recurrent Sun increase, the decreasing interval can be achieved by finding the grasp of the empty chain guide.

If the derivative is greater than (or greater than or equal to) 0, the function increases monotonically in that interval; The reverse is decreasing. Derivative of f(x), f(x).'=6x 2-2ax Discussion: When a=0, the function has no monotonically decreasing deficit interval; When a>0, let f(x).'

Solution: f(x)=x 3-3x 2+1

f'(x)=3x^2-6x

3x(x-2)

When x>2, f'(x)>0

When 0<=x<=2 f'(x)< 0

When x<0 f'(x)>0

Therefore, the single increment interval of the function is x<0 or x>2

The single minus interval is 0<=x<=2

f'(x)=x 2ax b, then f'(x) satisfies on interval [1,2]: f'(－1)=1－2a－b≤0；②f'(2)=4＋4a－b≤0。So your problem is to find the maximum value of the objective function z=a b over a region, which is a linear programming problem.

Solve this problem, you can.

y'=3x^2-6x+a=3(x-1)^2+a-3a>=3, y">=0, which increased monotonously in the entire banquet range.

a<3, x>mu reserve = 1+ (1-a 3) or x<1- (1-a 3) is a monotonic increase interval.

1-√(1-a/3

It's actually very simple, and I'm going to tell you about it, using formulas, or factoring, and you should know it.

The question is not wrong lz is not 3x, if it is the answer upstairs, if it is 9 is my answer).

f(x)=x³-ax+b-9

f'(x)=3x²-a

a 0 x = root number (a 3).

x — f at root number (A3).'(x) 0 f(x) single increment.

Root number (A 3) x root number (A 3) with f'Shirt disturbance (x) 0 (and not constant state Dan 0) f(x) single minus.

x — f at root number (A3).'(x) 0 f(x) single increment.

Single-increasing interval collapse (negative infinity, -root number (a 3)) root number (a 3), positive infinity) single decreasing interval - root number [(a 3), root number (a 3] a 0 f'(x) Evergrande at 0 f(x) on r.

a《When guessing 0, x is in negative infinity to 1-(a3) is x in 1-(a3) to 1+(a3).

x at 1-(a3) to positive infinity to

When a=0, x is neither nor when both 1 to positive infinity are 1 and 1 to positive infinity

a>0 x is in the range of real numbers