Let the function f x x 1 ax, where a 0

1) The original inequality is equivalent to ) (x +1) 1 +ax, squared on both sides.

a 2-1) x 2+2ax 0, and the two roots of equation (a 2-1) x 2+2ax = 0 are 0 and 2a (1-a 2).

So when a>1 2a (1-a 2) < 0, the solution set is .

When 00 and the solution set is {x 0x2>=0, the function f(x) is to be monotonic over the interval [0,+.

Then f(x1)-f(x2)<0 is immediately constant.

x1 2+1)- x2 2+1)]-a(x1-x2)<0 Heng.

deformed a>[ x1 2+1)- x2 2+1)] (x1-x2)

i.e. a>(x1+x2) [ (x1 2+1)+ x2 2+1)] (molecule is physicochemical).

Then use the inequality to deflect a>1

Because (x1+x2) [ (x1 2+1)+ x2 2+1)]Evergrande is in 1).

Solution: Because x 2+1 is greater than 0, f(x)=x 2-ax+11)f(x) is less than or equal to 1

x 2-ax is less than or equal to 0

Because a is greater than 0

So x belongs to (0,a).

2) f(x)=(x-a2) 2-a2 4+1 image symmetry axis is x=a2

To make (0, positive infinity) a monotonic function, because the opening is upward, it must be a monotonic increasing function.

So the axis of symmetry is on the y-axis or to its left side.

a 2 is less than or equal to 0

a is less than or equal to 0

Contradicts the title.

So there is no such a

The second question is not quite right. Try to write in a standard format.,But some of the symbols aren't very good at playing.,I'm sorry.

The derivative of f(x) is: [x ( (x +1))]a=[1 ( (1+(1 x) 2)]-a

Because 0<[1 ( (1+(1 x) 2)]<1, if it is monotonically decreasing, only 1<=a

If it is monotonically increasing, as long as a<=0

Summary. When x=0, the denominator is 0, and the function f(x) is undefined. Thus, the function f(x) does not exist at x=0.

function, f(x)=x xWhen x=0, f(x)=(aWhen x=0, the denominator is 0, the function f(x) is not defined. Thus, the function f(x) does not exist at x=0.

This is because in mathematics, the divisor cannot be 0, otherwise the operation is meaningless. In the definition domain of a function, there are points that make the denominator 0, so that these points do not belong to the definition domain of the function. For this problem, we can consider splitting x x into x and x x two parts.

When x ≠ 0, x x = 1, then f(x)=x x=x. When x=0, x x=0 0, the value of the formula is uncertain and cannot be directly substituted into the calculation. However, we can use the concept of limits to discuss the value of f(x) at x=0.

When x approaches 0, x x also approaches 0, so we can say that the limit of the function masking f(x) at x=0 is 0.

Proof: (Solution 1).

f(x)=|x+1/a|+|x-a|

a>0 when x>a

f(x) at 0

2x+1/a

a>x+1/a

a+1 a fundamental inequality).

While. -1/a2

basic inequality).

Solution 2) f(x)=|x+1/a|+|x-a|a>0

That is, find the number on the axis, the moving point.

x to two fixed points.

a,-1/a

.

That is, when x is between two fixed points, the sum of distances is the minimum, and the minimum value is the length of the line segment between the two fixed points.

So f(x).

f(x)min

a(-1/a)

(1) The inverse function of f(x) is p(x)=loga(x);

2) H(x)=loga(x-a)+p(x-3a)+g(x)), h(x)=loga(x-3a)+loga(x-a)=loga[(x-3a)(x-a)].

The domain is defined as (3a,+ is known by the condition, when x [a+2,+, h(x) is meaningful, so a+2>3a>0, i.e., 0 is established by h(x) 1 constantly, and f(x)=(x-3a)(x-a) 1 is constant, because f(x)=(x-3a)(x-a) has an image opening upwards, and the axis of symmetry is x=2a, so f(x)min=f(a+2)=2(2-2a) 1, and the solution is a 3 4.

Therefore, the range of values for a is [3, 4, 1].

Let h(x)=f(x)+g(x)=3x 2+1+x 3-9xh'(x)=6x+3x 2-9=3(x 2+2x-3)=3(x+3)(x-1)=0, and the extreme point x=-3, 1

x>1 or x<-3 monotonically increases.

3f(-3)=27+1-27+27=28 is the maximum.

f(1)=3+1+1-9=-4 is the minima.

Endpoint value f(2) = 12 + 1 + 8-18 = 3

From Kamiga, the maximum value of [k,2] is 28, then the Zen ridge k<=-3,

Answer: If the image of y=f(x) and the image of y=g(x) have two different common points a(x1,y1),b(x2,y2), that is, 1 x=ax +bx has and only two different solutions with muffle.

That is, ax +bx sedan -1=0 has and only two different solutions.

ax +bx -1=a(x-x1) (x-x2) i.e. ax +bx -1=a*[x -(2x1+x2)x +(x1 then line+2x1·x2)x -x1 ·x2].

x1 +2x1·x2=0,ax1 ·x2=1The option you gave is incorrect, please check and ask.

=x+a+a/x ，x>1

f'(x)=1-a/x^2=(x^2-a)/x^2a<1，x>1

Get f'(x)>0, f(x) in [1, is an increasing function.

2.Since the function is an increasing function within the defined domain, the original formula is equivalent to finding the inequality:

3m＞5-2m＞1

Easy to get m (1,2).

Let g(x)=g(x)+2x+3 2=x +(a+2)x+a+3 2 when x [2,5],g(x)>0, then only gmin>0 is required when the parabolic axis of symmetry x=-(a+2) 2 is at [2,5], then gmin=g(-(a+2) 2)=-(a+2) 2 4+a+3 2>0

Solution a When the parabolic axis of symmetry x=-(a+2) 2>=5, gmin=g(5)=25+5a+10+a+3 2>0 solves a

When the parabolic axis of symmetry x=-(a+2) 2<=2, gmin=g(2)=4+2a+4+a+3 2>0 is solved to a>-19 6

Let the function f(x)=2x 3-3(a+1)x 2+6ax+8, where a r, if f(x) obtains an extreme value at x=3, find the analytic expression of f(x)?

Answer: f'(x)=6x^2-6(a+1)x+6a=6(x-a)(x-1)

f(x) takes an extreme value at x=3, i.e.

f'(3)=6(3-a)*2=0

Yes a=3f(x)=2x 3-12x 2+18x+8 glad to help you. If you are satisfied, remember to "answer for satisfaction"! Happy you o( o