If the function has a monotonically decreasing interval, find the range of the value of the real num

A can be in the range of ( 1,+

Due to <>

Because the function f ( x ) has a monotonically decreasing interval, f'(x) <0 has a solution. And since the domain of the function is (0,+, then ax 2 +2 x 1>0 should have a solution of x >0. When a >0, y= ax 2 +2 x 1 is the parabola with the opening pointing upward.

ax 2 +2 x 1>0 always has a solution of x >0; When a <0, y= ax 2 +2 x 1 is the parabola with the opening pointing downward, and ax 2 +2 x 1>0 always has a solution of x >0, then =4+4 a >0, and the equation ax 2 +2 x 1=0 has at least one positive root. At this point, 1< a <0

When a=0; (0,1 2) decreasing, (1 2,+ increasing.

In summary, the value of a can be in the range of ( 1,+

Method: f ( x ) has a monotonically decreasing interval, so as long as f'(x) <0 is sufficient. In the classification discussion, carefully do each step.

<> "The function f(x) defines the domain (0,+ there is a monotonically decreasing interval f (x)=1 x-ax-2<0 in (0,+ there is a solution

When a 0, it is clear that f(x)=1 x-ax-2<0 has a solution in (0,+.

When a<0, so that f (x)=1 x-ax-2<0 is in (0, + has a solution, then a must be "-1

Therefore, the value range of a is (-1,+).

Answer] B Solution: The function f(x) has a monotonically increasing interval in the interval [1 2,2], and the function state deficit has a sub-quietly closed interval in the interval [1 2,2], so that the inequality f,(x)>0 holds. f,(x)=1/x+2(x-b)=2x2-2bx+1/x.

Let h(x)=2x2-2bx+1, then h(2)>0 or h(1 2)>0, i.e., 8-4b+1>0 or 1 2-b+1>0, to sum up, b<9 4

At that time, it was easy for us to obtain the analytic formula of the function, and then find the derivative function of the function, and discuss the symbols of the derivative function in a list to obtain the monotonic interval of the function; If the function is a subtractive function on the top, then it is constant on the above, which is transformed into the problem of the function is constant, and is transformed into an inequality that can be solved to obtain the range of values of the real numbers.

Solution: The domain of the function is defined as: At the time, .

When changing, the variation is as follows: - The minimum value can be seen from the above table, and the monotonically decreasing interval of the function is; The monotonically increasing interval is. The minimum is.

Points) by, get. If the function is a monotonically subtracting function on the top, then it is true at the upper constant, so the inequality is true at the upper constant. That is, it was established in Shangheng.

points) in turn is a subtraction function, so the minimum value is. So. (points).

The knowledge point examined in this question is to use derivatives to study the monotonicity of the number of socks, and the relationship between the monotonicity of the function and the derivative, in which the key to solving this question is to find the analytical formula of the derivative function according to the analytical formula of the number of keys of the original function.

According to the image and properties of the quadratic function, it can be obtained, and conclusions can be drawn.

Solution: The image of the quadratic function taken by the liquid potato is a parabola with an opening upward, and its axis of symmetry is, and the function is a monotonically increasing function in the interval, so yes, then the range of values of the real numbers is .

Therefore, the answer is to make trouble.

This question mainly examines the image and properties of quadratic functions, which is a basic question.

A can be in the range of ( 1,+

Since because the function f ( x ) has a monotonically decreasing interval, f'(x) <0 has a solution. And since the domain of the function is (0,+, then ax 2 +2 x 1>0 should have a solution of x >0. When a >0, y= ax 2 +2 x 1 is the parabola with the opening upward, and ax 2 +2 x 1>0 always has a solution of x >0; When a <0, y= ax 2 +2 x 1 is the parabola with the opening pointing downward, and ax 2 +2 x 1>0 always has a solution of x >0, then =4+4 a >0, and the equation ax 2 +2 x 1=0 has at least one positive root.

At this point, 1< a <0

When a=0; (0,1 2) decreasing, (1 2,+ increasing.

In summary, the value of a can be in the range of ( 1,+

Method: f ( x ) has a monotonically decreasing interval, so as long as f'(x) <0 is sufficient. In the classification discussion, carefully do each step.

Known <>

Function <>

In the interval [<>

On a monotonically decreasing, then the real number <>

The value range is ( )a [<

b．(<

c．[<

d．(0,2]

C. Question Analysis: When <>

<> monotonically decreasing, so <>

So.

So.

Comments: This question examines the monotonicity of trigonometric functions, and the key to solving the problem is to be able to solve the problem by combining the monotonicity of the sine function with the overall idea, which is a mid-range question.

Put the substitution, solve the inequality, you can;

On the monotonically decreasing, that is, it is held in the upper constant, and the range of values that can be solved by classification and discussion.

Solution: , the domain is defined as.

The monotonically increasing interval of OR , BY, AND IS , and the monotonically decreasing interval is .

Cause. At that time, the function was monotonically decreasing on it.

At that time, it was a quadratic function, and its axis of symmetry was, if and only if, that is, when the wheel contained, at this time there was no solution.

At that time, it was a quadratic function, if and only immediately. , the function is monotonically decreasing on it.

To sum up, the range of values for real numbers is.

This question examines the relationship between derivatives and the monotonicity of functions, which for derivables (not always) are sufficient and necessary conditions for monotonically decreasing intervals.