Whether a exists such that the function f x x 3 3 a 1 2 x 2 3ax is monotonic

f(x)=x -[3(a+1) 2]·x +3axf (x)=3x -3(a+1)x+3a, the image opening of the derivative function is upward.

Derivative discriminant = [3(a+1)] 12a=9(a-1 3a+1)=9(a-1 6) +35 4 Evergrande is zero.

No matter what value a takes, f (x) always takes a negative value.

That is, there is no such thing as a making f(x) a monotonic function.

=(3a-2) 2-4(a-1)=9(a-8 9) 2+8 9>0If there is a real number a that satisfies the condition, then only f(-1)*f(3) 0 is sufficient.

That is, f(-1)*f(3)=(1-3a+2+a-1)*(9+9a-6+a-1).

4(1-a)(5a+1)≤0

A -1 5 or A 1

Test: When f(-1)=0, a=1

f(x)=x^2+x.Let f(x)=0, i.e. x2+x=0

Get x=0 or x=-1

The equation has two roots on [-1,3], which is not in line with the topic, so a≠1 when f(3)=0, a=-1 5

At this time, f(x)=0, that is, x 2-(13 5)x-6 5, so that f(x)=0, that is, x 2-(13 5)x-6 5=0

The solution yields x=-2 5 or x=3

The equation has two roots on [-1,3], which is not in line with the topic, so a≠-1 5

In summary, the value range of a is (- 1 5) (1,+

Let f(x) = f(x)-f(1 2).

x^3+x^2+ax+1)-(1/8+1/4+a/2+1)

x^3+x^2+ax-3/8-a/2)

Finding the solution of the equation f(x) = f(1 2) in addition to 1 2 on (0,1) is equivalent to finding the zero point of f(x) on (0,1) in addition to 1 2.

f'(x)=3x^2+2x+a=3(x+1/3)^2+a-1/3

Order f'(x)=0 is easy to solve to obtain the two extreme points of f(x) as .

x1=-1/3*[1+√(1-3a)]，x2=-1/3*[1-√(1-3a)]

It is easy to know that x1 is always a negative number, which is the maximum point.

When x2 0 or x2 1, f(x) is a monotonic function on (0,1) and cannot have more than one zero.

To have another zero point, there must be 0 f(1 2)=0 constant, and f(0)=-3 8-a 2, f(1)=13 8+a 2

Since f(x2) is the minimum, the endpoint value must be greater than or equal to 0

i.e. f(0)=-3 8-a 2 0 (2).

and f(1)=13 8+a 2 0 (3).

Lianli (1), (2), (3), can be solved.

5 takes the intersection to get a [-13 4,-7 4) (7 4,-3 4].

When the value range of a is [-13 4,-7 4) (7 4,-3 4], there is x (0,1 2) (1 2,1) such that f(x) = 0, i.e. f(x) = f(1 2).

Solution:( Since f(x)=3x2+3(1-a)x-3a=3(x+1)(x-a), and a 0, f(x) decreases monotonically on [0,a] and increases monotonically on [a,+

f (0) = 1, f (a) = -12a3-32a2 + 1 = 12 (1-a) (a + 2) 2-1

When f(a) -1, p=a

In this case, -1 f (x) 1 holds when x [0,p].

When f (a) -1, since f (0)+1=2 0, f (a)+1 0, there is p (0,a) such that f(p)+1=0

In this case, -1 f (x) 1 holds when x [0,p].

In summary, for the positive number a, there is a positive number p, such that when x [0,p], there is -1 f (x) 1

7 points) ) From ( ) we know that the minimum value of f (x) on [0,+ is f (a).

When 0 a 1 and f (a) -1, then g(a) is the real root of the equation f (p) = 1 satisfying p a, i.e., 2p2 + 3 (1-a) p-6a = 0 satisfying the real root of p a, so.

g（a）=3(a-1)+

9a2+30a+94．

and g(a) increases monotonically on (0,1], thus.

g（a）max=g（1）=3．

When a 1, f (a) -1

Since f (0) = 1 and f (1) = 92 (1-a)-1 -1, therefore.

0，p]⊂[0，1]．

At this point, g(a) 1

In summary, the maximum value of g(a) is 3

Hmm, you can take a look at the original!

Find the derivative and judge the monotonicity of the function.

^f'(x)=3x^2+2(1-a)x-a(a+2)

Because the function f(x) is not monotonic over the interval (-1,1), it is on the interval (-1,1).'(x) It is not possible to maintain a constant symbol.

f'The image of (x) is a parabola with an opening upward, and to meet the above conditions, it is necessary to have: f'The image of (x) has two intersections with the x-axis, and at least one of the intersections is within the interval (-1,1).

f'(x)=(x-a)(3x+a+2), let f'(x)=0, get two: x1=a, x2= -(a+2) 3

So, there should be: a≠ -a+2) 3

And -1 gets: a≠ -1 2 and -1 so when a (-1,-1 2) (1 2,1) satisfies the condition.

In the case that the two roots are equal to the boundary values respectively, i.e., x1=1 (or -1) and x2=1 (or -1), the calculation shows that they do not meet the requirements.

In summary, a (-1,-1 2) (1 2,1).

Derivative f'(x)=3x^2+2(1-a)x-a(a+2)x1=[(a-1)-|2a+1|]/3 x2=[(a-1)+|2a+1|]/3

When 2a+1 0, i.e. a -1 2, x1= - a+2) 3 x2=a

x1<-1 and x2>1. Get a>1

x1<-1 and x2>1. get a -1, a>1 contradiction x1<-1 and x2 1A>1, a 1 contradiction, when 2a+1<0, i.e., a<-1 2, x1=a, x2= - a+2) 3

x1<-1 and x2>1. Get a< -5

x1<-1 and x2>1. Get a<-5, a -1 contradiction x1<-1 and x2 1Get -5 a< -1 in summary: a (-1) (1,+.)

1. Analysis: It is difficult to directly find the value range of a when it is not monotonic, so it may be simpler from the opposite side: that is, to find the value range of a when the function is monotonic within a given defined domain, which is transformed into a problem of analyzing the derivative function of f(x);

2. Derivative of the function f(x), f'(x)=3x^2+2(1-a)x-a(a+2)；(Delta = 16a 2 + 16a + 4, Evergrande is equal to 0).

3. To make a function monotonous, you need to make f'(x) satisfies that in a given defined domain, the evergrandity is equal to 0 or constant less than or equal to 0, so that the problem is transformed into a problem of the position relationship between the unary quadratic equation and the x-axis; [The two roots of the equation are x1=a and x2=-(a+2) 3].

Fourth, when f'(x) When Evergrande is equal to 0 in a given defined domain, when delta=0, i.e., a=-1 2, the condition is satisfied;

When delta is greater than 0, i.e., a is greater than -1 2, x1>-1 2, x2<-1 2, and the position of the x-axis is related by the unary quadratic equation.

Conclude that the conditions are not met;

When f'(x) When it is always less than or equal to 0 in a given defined domain, the range of values of a can be found similarly according to the above method (because you don't have a pen and paper in your hand, here you are, you should know how to do it below).

According to the inscription f(x) [at least] has an extreme point in the interval (-1,1), due to f'(x)=3x 2+2(1-a)x-a(a+2)=(x-a)[3x+(a+2)],a≠-1 2, f(x) has two different extreme points x1=a and x2=-(a+2) 3,a=-1 2, f(x) strictly monotonically increases.

1 i.e. -1 -1

f'(x)=3x²+2(1-a)x-a(a+2)=(x-a)[3x+(a+2)]

If a=-(a+2) 3, a=-1 2

then f'(x)=3(x-1/2)²>=0

At this point, it's a monotonic function on r, which doesn't fit the topic.

a≠-1/2

f'(x)=0 has two unequal roots.

In (-1,1) is not monotonous.

That is, there are increasing functions and subtracting functions.

So the derivatives are positive and negative in this range.

So f'The root of (x)=0 is in this range.

f'(x)=(x-a)[3x+(a+2)] The two roots are x=a, x=-(a+2) 3

Then -1-3-5 is combined with -5

The third approach is correct. And the best.

Solution: There are two cases where the function is known to have a maximum value in the given interval:

1. There is a certain number of 0 versions monotonous on the interval [x0,1].

Weights. i.e. the derivative "0,, and f(1)=1, i.e., a=1, when a=1, find the derivative, f'=3x 2+a 3x 2+1>0 So at this time a=1 satisfies the condition.

2. Only the maximum value is obtained on the given interval, that is, there is a certain number of 0f'=3x^2+a

Order f'=0 3x 2+a=0 a<0, x= (a 3) because 0f( (a 3)=(a 3) (3 2)+(a 3) (3 2)=1a=(1 4) (1 3) This contradicts a<0, so there is no solution at this time.

The composite has a = 1

Solution: (1) Let f'(x)=ax 2-2(a+1)x+4>0, then the bridge (ax-2)(x-2)>=0

When a=0, there is x>=2, and the monotonic increase interval is [2,+

When a>0, there is (x-2 a)(x-2)>=0If 0=2 a or x<=2, then the monotonic increase interval is [2 a,+ and (- 2];

If a=1, then the inequality is constant, and the monotonic interval of the liquid stool is (-

If a>1, then x<=2 a or x>=2, the monotonic increase interval is [2,+ and (- 2 a];

When a>0, there is (x-2 a)(x-2)>=02 a<=x<=2, that is, the monotonic increase interval is [2 a,2].

2) a<0, then the function decreases on (- 2 a) and increases on [2 a,0].

When a>=-2, 2 a<=-1, i.e., the function is increasing on [-1,0], then the function has a minimum value f(-1)=-a 3-(a+1)-4+1=-4a 3-4=-3, a=-3 4

When a<=-2, 2 a>=-1, i.e. the function decreases on [-1,2 a] and increases on [2 a,0]. Then the function has a minimum value f(2 a) = (3a 2+12a-4) (3a 2) = -3, a = -1 2-(7 12) (1 2), inconsistencies and assumptions, rounded off.

In summary, there is a negative real number a=-3 4, such that x [-1,0] and the function has a minimum value of -3

f(x)=1 3x complex 3-1 2ax 2+(a-1)x+1

To confirm, above.

Whether that function is of this form.

f(x)=(1/3)x^3-(1/2)ax^2+(a-1)x+1

Yes, do it.

First, the derivative of f(x) is obtained, and g(x)=x 2-ax+a-1 is obtained

Think of the above function as a function of a, yes.

q(a)=(1-x)a+x^2-1

Since it is a subtractive function in the interval (1,4), (1-x)a+x 2-1<0 (1).

In the interval (6, is the increasing function, so (1-x)a+x 2-1>0 (2).

Solve inequalities (1), (1-x)a<1-x 2

a>(1-x 2) (1-x) Note: 1-x is less than 0, so the inequality should be changed.

So a>1+x, x is greater than 1 and less than 4, so that the equation always holds, a>1+4, i.e., a>5.

For example, if x takes 2, then a > 3, but when x takes 3, the range of a may not be true. ）

Solve inequalities (2), (1-x)a>1-x 2

a>(1-x 2) (1-x) Note: 1-x is less than 0, so the inequality should be changed.

So a<1+x, x is greater than 6, so that the equation always holds, so a<7