The known functions f x 2ax 1 3 3 1, x 0 1, a R

1. When a=1, f(x)=2x-(1 3 3)+1, because x(0,1], then f(1)=3-(1 3 3)>2 Therefore, the image of the function f(x) is not always below the line y=2.

2. When f(x) is an increasing function on x (0,1], that is, for any 0f(x1)f(x2)-f(x1)=2a(x2-x1), because x2-x1 >0, so to f(x2)-f(x1)>0, it must be a>0, so the value range of a is (0, positive infinity).

In some places, the increase function refers to the function is not reduced, so it is not to seek "is to seek" =, I don't know what the landlord requires, if it is "=", just add an equal sign to the "number", and then the value range is [0, positive infinity))).

3. From the point of view of f(x2)-f(x1)=2a(x2-x1) in question 2, if a>0, f(x) increases monotonically, if a<0, f(x) decreases monotonically, if a=0, f(x) is a constant function, so f(1) is the maximum value when a>0, f(1)=3-(1 3 3).

When a<0, f(0) is the maximum, f(0)=1-(1 3 3) When a=0, f(x) is equal everywhere, and any point is both the maximum and the minimum, f(x)=1-(1 3 3).

No, because in this interval the maximum value is greater than 2, the minimum value is less than, and the derivative is greater than zero.

The piecewise function swims the sail, and the segmented solution is open.

For reference, please smile and accept.

Monotonic increment: [-1,0)u[2,5].

Range: [-1,3].

Solution set: -1 x 5 and x ≠2

Solution: (1) From the meaning of the title, it can be known:

When x [-1,2], f(x)=-x2+3, is part of the quadratic function;

When x (2,5], f(x)=x-3, is part of the primary function;

2) From the image of the function, it can be seen:

The monotonically increasing intervals of the function f(x) are: [-1,0] and [2,5] As can be seen from the graph, the solution set of f(x) 1 is: -1 x 2 or 4 x 5

I think so.

Considering the situation of the three-shade Zen type, 1) 01, just bring it in directly.

3) When m=1, there is no result, because the denominator cannot be 0

（1）：f(x)=x^3-3ax^2+1，f'(x)=3x 2-6ax, let f'(x)=0，x1=0，x2=，f(x2)=1-4a^3。

2): If y=f(x) has three different zeros, then y=f(x) has two different extremums, i.e., x1≠x2,a≠0. and f(x1)*f(x2)<0,1-4a 3<0,a>4 (-1 3).

According to the curve of the unary cubic function, because f(x1) >0 and f(x2) <0, x2> x1, a>0. In summary, a>4 (-1 3).

f'(x)=(1/4)x^2+(a+1)x+4a+1f'(x) is a quadratic function of the opening upwards, and it is impossible to have an constant f on r'(x) <=0, only the spike hand can be f'(x) > comic = 0

i.e. f'Guess that (x) and x-axis have at most one common point.

Discriminant formula = (a+1) 2-(4a+1) = a 2-2a<=0 Therefore, the range of values for a is [0,2].

Because f(x) = (1 12) x 3+[(a+1) 2] x 2+(4a+1)x

So f'(x) = (1 4) x 2 + (a + 1) x + (4a + 1) because g(x) = f'(x) is an even function.

So a=-1

So f'(x)=（1/4)x^2-3

Let g(x)=f'(x) = (1 4) x 2-3 = 0 to get x = 2 3

So the maximum value of f(x) is f(-2 3)=4 3 and the minimum of f(x) is f(2 3)=-4 3