A question about the monotonicity of a higher function

1) f(x)=x*2+2ax+2,x [-5,5] is a part of the quadratic function f(x)=x*2+2ax+2,x r image, as long as f(x)=x*2+2ax+2,x [-5,5] is a monotononic function on one side of the vertex of the quadratic function f(x)=x*2+2ax+2,x r.

That is, as long as the vertices of this quadratic function are not in the [-5,5] interval, the function can be monotonic.

Let the vertex x-coordinate 5 or -5 solve: a -5 or a 5 when a -5 or a 5.

The function f(x)=x*2+2ax+2 x [-5,5] is a monotonic function.

2)f(x)=(x^2+2x+a)/x=x+a/x +2f(x)>0

x+a/x>-2

When a>=0.

f(x) is the hook function, the minimum value is x= a, i.e., 2, a >-2, because a >0, so a [0, positive infinity) is true, and when a<0.

f(x) is an increment function when the minimum value is x=1.

1+a>-2

So a>-3 So a (-3,0).

So to sum up a (-3, positive infinity).

or because f(x)=(x 2+2x+a) x,x [1, positive infinity)f(x)>0

x 2+2x+a>0.

x+1)^+a-1>0

At this point, the function can be set up immediately when it meets the minimum of x.

x=1 at 4+a-1>0

a>-3

3) Solution: y=x 2-2ax-1

x-a)^2-1-a^2

axis of symmetry x=a

If a<0 then at x>=0 is monotonically increasing, so the minimum value is f(0)=-1 and the maximum value is f(2)=4-4a-1=3-4a

If 0<=a<=2 then on 0<=x<=2, the minimum value is f(a)=-1-a2 and the maximum value is max=max

If a>2 then subtract from 0<=x<=2, so the minimum value is f(2)=3-4a, and the maximum value is f(0)=-1

or a 5

It is to look at the position of the axis of symmetry, and it is required to be monotonic in the interval, then a is not (5, 5).

In this problem, if x belongs to [1,+8)f(x)>0 constantly, then there is no intersection within the interval range, and 0 can be solved.

3。The axis of symmetry is a, the opening is upward, when a 0, the minimum value is f(0), and the maximum value is f(2); When 0 a 1, the minimum value is f(a) and the maximum value is f(2); When 1 a 2, the minimum value is f(a) and the maximum value is f(0); When a 2, the minimum value is f(2) and the maximum value is f(0).

This function is a subtractive function, which means that as x increases the track and the accompanying rock increases, the corresponding y value gradually decreases. Mess up.

So in f(x)>f(2-x), the y value on the left is larger than the y value on the right, which means that the x value on the left is smaller than the x value on the right.

This gives x<2-x.

Solution: Because the function f(x)=a|x-b|On (2,+ is a subtraction function, so a is not equal to 0, this function is a one-time function.

1.If x>b, then f(x)=ax-ab

So a<0

2.If x03If x=b, then f(x)=0, which does not fit the title, so x is not equal to b.

Because x belongs to (2,+ is a subtraction function, so b is less than or equal to 2,x>b then f(x)=ax-ab, so a<0

In summary: b

Dizzy.,It's better to draw a picture of this question.,You can see it at a glance.,How can I draw it for you.。。。 Anyway, a<0, b<=2

Using the principle of increase and subtraction to satisfy the subtraction function, when a is greater than 0, then |. In (2,+ should be a subtraction function, i.e. x increases by |x-b|If you decrease it, then b must be greater than 0

When a is less than 0, then the same is true|. If it should be an increment function, then x increases by |x-b|If it also increases, b must be less than or equal to 2 and it is done, and I don't know how to contact it.

This is a one-time function, which is either a subtraction, an increase, or a constant function, and it is easy to know that a1<0 b is a real number.

If you don't know how to find a derivative, use the definition: for any x2 and x1, where 0 root number (x1 +1) - root number (x2 +1) = (x1 + x2) (x1 - x2) (root number (x1 +1) + root number (x2 +1)) is actually the square difference formula.

The common factor (x1+x2) is proposed, and then x1 is used to use x1 "root number (x1 +1), x2 "root number x2 "root number (x2 +1)".

Get (x1+x2) (root number (x1 +1) + root number (x2 +1)) <1 typing is tired, give points!

Try finding the derivative greater than zero or combine the root number operation and the proof of the monotonicity of the curve region of the quadratic equation, which should be proved quickly.

1.Find the derivative of the function, f'(x)=2*x 2*(root(x+1)) a = x(root(x+1)) 1

Because x (root number (x +1)) is always less than 1, f on [0,+'(x) <0 is constant, i.e. the function f(x) is monotonic on [0,+.

Subtract the function. 2.If you haven't learned the derivative, you can let x=tanx, where x belongs to [0,90 degrees], then there is.

f(x)= 1/ cosx - a * tanx

Let x1>x2>=0, then there is f(x1)-f(x2)=1 cosx1-atanx1-(1 cosx2-atanx2).

1/cosx1-1/cosx2-a*(tanx1-tanx2)

cosx2*(1-asinx1)-cosx1*(1-asinx2) )/cosx1*cosx2

( cosx2*(1-sinx1)-cosx1*(1-sinx2) )/cosx1*cosx2

Because x1>x2, cosx1sinx2. The above numerator <0, denominator》0, so there are f(x1)-f(x2)<0

That is, when x1>x2>=0, f(x1)-f(x2)<0.

So the function f(x) is a monotonically subtracting function on [0,+.

Take x1, x2 on [0, positive infinity) and set x1 > x2.

y1-y22(x1)^4-2(x1)^4

2(x1^4-x2^4)

2(x1 2+x2 2)(x1 2-x2 2)=2(x1 2+x2 2)(x1+x2)(x1-x2).

The value of the first and second parentheses is greater than zero, and the value of the third parenthesis is also greater than zero because x1>x2. So y1-y2>0, i.e. the function y=2x 4 is incremented within the defined domain.

a>b>=0

a^4-b^4=(a-b)*(a+b)(a^2+b^2)>0

By definition, the function y=2x to the power of 4 (4 in the upper right corner of x) is increasing on [0, positive infinity].

Solution: According to the meaning of the topic.

1<=x-2<=1

1<=1-x<=1

1-x<=x-2

Get: 1<=x<=3

0<=x<=2

x>=3/2

To sum up: 3 2<=x<=2 (if it is strictly incrementing, change the left side to less than).

Method 1: Derivative method.

Seek a derivative of f, and know f'=x channel (x 2-1) when x > 1

time, x 2-1>0

So f'>0

So f(x) is monotonically incremented on (1, positive infinity).

Method 2: Definition.

x1>x2>1

f(x2)-f(x1)

x2^2-1)-√x1^2-1)

(x2^2-1)+√x1^2-1))/x2^2-1)-√x1^2-1))*x2^2-1)+√x1^2-1))

(x2 2-1)+ x1 2-1)) x2 2-x1 2) >0 can also belong to the same to prove to be an increasing function.