f x 3ax 2 bx is an even function defined on a 1,2a, discussing the monotonicity of g x f x 2 x

Because f(x) is an even function on [a-1,2a], f(2a)=f(-2a) yields: 12a 3+2ab=12a 3-2ab 4ab=0

When a=0, the domain is defined as: [-1,2] does not contradict the y-axis symmetry with respect to the topic even function.

So b=0 so f(x)=3ax 2 defines the domain symmetrically with respect to the y-axis, so a-1=-2a solves as: a=1 3 so f(x)=x 2

g（x）=x^2+2/x

Derivative: g(x).'=2x-2/x^2=(2x^3-2)/x^2=2(x-1)(x^2+x+1)/x^2=2（x-1）[(x+1/2)^2+3/4]/x^2

So g(x) at (1,+infinity) is an increasing function and g(x) at (0,1] (infinity,0) is a decreasing function.

According to the even function, it is defined as a-1=-2a and 3ax 2+bx=3ax 2-bx

So a=1 3, b=0

So f(x)=x 2

So g(x)=x 2+2 x=(x 3+2)*1 x because the function x 3+2 is increasing on r, and the function 1 x is decreasing at (negative infinity, 0) and (0, positive infinity).

So f(x) decreases in (negative infinity, 0) and (0, positive infinity).

Solution: f(x)=3ax 2+bx is an even function defined on [a-1,2a].

So b=0, a-1=-2a, then a=1 3

Therefore, g(x)=f(x)+2 x=x 2+2 x (x is not equal to 0)g'(x)=2x-2 x 2=2(x 3-1) x 2 which when x>1, g'(x) >0 increase.

When 0, the monotonically increasing interval of g(x) is [1, positive infinity).

The monotonically decreasing intervals are (negative infinity, 0) and (0, 1).

From f(x) is an even function, we know that b=0 and a-1+2a=0, i.e., a=1 3, then g(x)=x 2+2 x

Next, find the derivative, and find the extreme point as 1

Then increase interval: [1,+ infinity), subtract interval: (-infinity,0),(0,1].

The axis of symmetry equation of the function f(x)=x2-2ax+3 is exemplified as: x=a, and when a -2, the function f(x)=x2-2ax+3 increases monotonically between the trillions of rents in the district (-2,2). When -2 a 2, the function f(x)=x2-2ax+3 decreases monotonically in the interval (-2,a) and increases monotonically in the interval [a,2); When a2, the function f(x).

（1）f’（x）=4x³+3ax²+4x

When x=-10 3

f’（x）=4x³-10x²+4x

Let f'(x) 0

Function increments. It is solved by the needle and thread method.

0 x 1 2 or x 2

So the increment interval is [0,1 2] and [2,+

Similarly, the subtraction intervals are (- 0) and (1 2, 2).

2) The function f(x) has an extreme value only at x=0, indicating that f'(x)=0 has only one solution cut to 0

f'(x)=4x +3ax +4x=x(4x +3ax+4) satisfies only one solution cut of 0

Only 4x +3ax + 4 = 0 has no solution.

9a²-4*4*4＜0

The -8 3 function f(x) = x 4 + ax 3 + 2x 2 + b (x r), where a, b rIf the inequality f(x) 1 is constant on [-1,1] for any a [-2,2], the range of values for b is obtained.

y'=4x^3+3ax^2+4x=x(4x^2+3ax+4)δ＝9a^2-64<0

y"=12x^2+6ax+4

36(a^2-16/3)<0

Obviously the function f(x)=x 4+ax 3+2x 2+b(x r), where a, b r, has only one maximum value on ( and is concave over the entire interval.

There is according to the title.

max f(x)=max=max=5+b

And because for arbitrary a [-2,2], the inequality f(x) 1 is constant on [-1,1].

So b -4

If the function f(x)=ax3+3x2-x, discuss the monotonicity of f(x)? Solution: From the problem to know a≠0, f (x) = 3ax2-6x= , let f (x) = 0, get x1 = 0, when a 0, if x (-0), then f (x) 0, so f(x) is an increasing function on the interval (- 0);

If , then f(x) 0, so f(x) is a subtractive function in the interval ;

If , then f(x) 0, so f(x) is in the interval , and is an increasing function on .

When a 0, then f(x) 0, so f(x) is a subtraction function over the interval ;

If , then f(x) 0, so f(x) is an increasing function over the interval , and if x (0,+ then f(x) 0;

So f(x) is a subtraction function over the interval (0,+.

3. Combing of test sites.

The monotony of the function.

The relationship between the derivative and the monotonicity of the function:

1) If f (x)>0 is constant on (a,b), then f(x) is an increasing function on (a,b), and the corresponding interval of the intersection of the solution set of f (x)>0 and the defined domain is the increasing interval;

2) If f (x)<0 is constant on (a,b), then f(x) is a subtraction function on (a,b), and the corresponding interval of the intersection of the solution set of f(x)<0 and the defined domain is the subtraction interval.

A general procedure to solve the monotonicity of a polynomial function using derivatives:

determine the domain of definition of f(x);

calculate the derivative f (x);

Find the root of f(x)=0;

The definition domain of f(x) is divided into several intervals with the root of f(x)=0, and the symbols of f(x) in these intervals are examined listically, and then the monotonic interval of f(x) is determined: f(x) >0, then f(x) is an increasing function in the corresponding interval, and the corresponding interval is an increasing interval; f(x)<0, then f(x) is a subtraction function in the corresponding interval, and the corresponding interval is the subtraction interval.

If there are finite points in an interval such that f (x) = 0, and there is constant f (x) > 0 in the rest of the points, then f(x) is still an increasing function (the situation of the subtracting function is exactly the same), that is, in the interval f (x)>0 is a sufficient condition for f(x) to be an increasing function in this interval, but not a necessary condition.

The known function f(x)=x ex, g(x)=-x2-2x+m (1) finds the monotonic interval of the function f(x); （.

Answer. This question is worth 13 points) known functions. (1) When and , when the sub-expression containing is tried, and.

Answer. It is known that f(x)=x2+ax+a(a 2,x r), g(x)=e-x, (x)=f(x) g(x) (1) when a...

Answer. out of 12 points) set function. ( If there is a domain in the definition that makes the inequality ,...)

Answer. This question has a total of 14 points) is a known function, and it is an odd function ( find the value of ,; ( Finding the function.

Answer. Known functions, where. The tangent equation at the point is , then the functions a=, b=

This is a simple math problem, probably to learn this kind of function in high school, so if you want to find the unknown first, first you have to exchange all the unknowns into an unknown, so that you can calculate it better to be able to calculate its monotonicity.

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

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

Therefore, the spine filial piety is b=0;

f(2)=5/3, (2a+2)/(3*2)=5/3, 2a+2=10, a=2;

So f(x)= 2x+2) (3x),2) is an increment function on (- 1).

Let any x1, x2 (-1), and x11, -x2>1, (x1)(-x2)>1, x1x2>1>0

x1x2-1>0

So, f(x1)-f(x2)<0, i.e., f(x1).< f(x2), so the conclusion holds.

f(-x)＝f(x)

i.e. (a-1) (-x) 2-2a(-x)+3 (a-1) x 2-2ax+3

then a 0f(x) -x 2+3

Therefore, the function increases monotonically in [- 3].

Solution: f(x)=2x

3/3+x2+ax+

b，x>

1, find the derivative f'(x)=2x

2+2x+a, so that f'(x)=2x

2+2x+a=0=>△=4

8a<

0, so for everything x

1 has f'(x)=2x

2+2x+a Evergrande is 0, so f(x) is in x

1. Monotonically incrementing.

Solution: (1) f'(x)=x^2-2(1+a)x+4a=(x-2)(x-2a)

a>1,2a>2

x<2,x>2a,f'(x) >0 and increment the function by 20.

f(0)=24a>=0,a>0

f(2a)=(-4/3)a^3+4a^2+24a>0a^3-3a^2-18a<0

a(a-6)(a+3)<0

a<-3,0 In summary:

1