50 A high school function discriminant problem

The title is not clear, it seems that it has not been finished, can you explain it again? What is the relationship between m0 and m?

I think you're thinking too complicated about the problem, the above function requires that any x is true, and the following function f(x) = 2 (sinx+cosx) does not satisfy the condition at 0, so f(x) is a function that does not meet the above conditions.

As for the slope you are talking about, it must not be, because the slope at a point is the derivative of that point, and the definition of the derivative at x0 is.

f'(x0)=lim(x->x0)[f(x)-f(x0)] (x-x0), so f(x) x does not indicate slope.

If f(x) x has the limit k when x approaches infinity, he may have an oblique asymptote with a slope of k. But whether there are known conditions is not enough.

This is the definition of a bounded function. And because the domain of the function f(x) is r.

f(x) = 2(sinx+cosx) is not in accordance with the nature of this function. Because) = 2(sinx+cosx)<=2sin(x+pi 4)<=,|x|< = m2, i.e. -m2< = x< = m2

When x is not equal to 0, (f(x) x does not represent the slope, then this (sinusoidal) slope is not positive infinity.

For example, f(x)=x,|f(x)|=|x|<=m|x|, so m>=1 I don't know if you understand? ）

Because the opening of the function is upward, in the negative infinity to 1 monotonic, it is monotonically reduced, and the minimum value is obtained when x is equal to 1, as long as the minimum value is greater than 0, the function has a solution, because the true number must be greater than 0.

Because the original opening is upward, and at negative infinity to 1 is a monotonic function, so the monotonic decreases, and the lowest point should be to the right of 1.

The lowest point is x = a > = 1

Again, original 0

So the minimum value of 0, because the function is monotonically decreasing from negative infinity to 1, is substituted to obtain.

1-2a+3>0

A modern definition of a function.

is given a set of numbers a, assuming that the element in it is x, and applying the corresponding rule f to the element x in a, denoted as f(x), to get another set of numbers b, assuming that the element in b is y, then the equivalence relationship between y and x can be expressed by y=f(x), and the concept of function has three elements: the domain a, the domain b, and the corresponding law f. The core of this is the correspondence law f, which is the essential feature of functional relations.

The parameters of this quadratic function a=1,b=-2a,c=3Since the function is monotonic on (- 1], the vertex of this quadratic function is to the right of x=1 (can be taken) [vertex 1].

Whereas the vertices of the quadratic function are-b/2a=-(2a)/2*1=a

That is, a 1Since a=1>0, the function opening is upward. The simultaneous function definition domain contains (-1], which means that in this range, the true numbers are all greater than 0, and at the same time, the quadratic function is partially monotonically decreasing, and the minimum value of the quadratic function is x=1

Therefore, the minimum value must be greater than 0, i.e., 1 -2a*1+3>01-2a+3>0

Drawing method. Because the original opening is upward, and it is a monotonic function to 1 at negative infinity, it is monotonically decreasing, and the lowest point should be to the right of 1, and the lowest point is x = a > = 1

Again, the original formula 0, so the minimum value of 0, because the function is monotonically decreasing from negative infinity to 1, is substituted to obtain.

1-2a+3>0

Because this function is monotonic, and the coefficient before x is greater than zero, it can only decrease monotonically in this range, and the axis of symmetry must be on the right side of the range, so a 1, we know that the function is decreasing from the front, so when x takes one, it must be the minimum value, and as long as this value is greater than zero, then all function values are greater than zero, so these two equations are obtained.

True numbers are quadratic functions, monotonicity is related to the axis of symmetry, according to the image, the axis of symmetry should be greater than or equal to 1, as a true number must be meaningful, when x = 1, the true value is greater than 0.

You're wrong, because when the second car goes, the car behind is also going, and when the second car is at the end, the third car is only KV S away, and in the same way, the car behind is the same, ten cars, there are 9 gaps, so it's +9kv, if you still can't figure it out, you can think of them as a train.

Haha, the second car etc., the first car kv 2 v = kv, the third car etc the second car kv 2 v = kv

.The 10th car and the 9th car KV 2 V = KV The total time interval is 9KV

That is, the earliest car such as the twelfth car waited for a total of 9kv time, as for how much time the middle car waited, you don't have to repeat the accumulation, understand the deviation.

You can think of these wagons as a train, and the length of the train can be known.

In the end, what you want can be seen as the time it takes from the start to the end of the train to the station.

You are wrong in calculating the length of the car, which is equal to the total length of the 9-car box, which is the sum of the interval lengths in the question.

I think of it this way: if you start at the same time and ask for 2000 v s, because of the interval, I want to wait for the second car and wait for the first car kv 2 v = kv s, so that the third car does not have to wait for 2kv

You're right to think about this, and keep thinking along the lines.

The distance between the fourth car is 3kv, and the fifth car is 4kv ,..Then the last car is, of course, 9kv

You didn't think wrong, you misunderstood.

The distance from the first car to the departure to the last car should be 2000+k*v 2*9, then the time should be divided by v, that is, 2000 v+9kv

The time should be calculated with t=s v, the third car is immediately followed by the second car, and the distance is also kv 2, so the time of waiting for the second car kv, so the time of waiting for the first car is 2kv, and it is 9kv in turn, you know.

"If you start at the same time to 2000 v s, because you want to interval, I want to be the second car and wait for the first car kv 2 v = kv s, so that the third car is not waiting for 2kv", here, you are right, but how can it be 45kv in the back? According to your above idea, the first car and the fourth car have to wait for 3kv, and the first car and the fifth car have to wait for 4kv... The first car and the tenth car have to wait for 9kv, plus the time for the 10th car to run 2000m, it should be 2000 v+9kv, the 45kv you get, I guess you should put 1kv, 2kv, 3kv...

9kv adds up, right? That's wrong, it's impossible to go back to the starting point and wait for the second car after the first car waits for the second car!?

Because I want to be spaced, I want to be the second car and wait for the first car kv 2 v = kv s, so that the third car is not to wait for 2kv so that the answer is 2000 v + 45kv The answer is +9kv, I ** think wrong hungry?

I wrote it wrong, the third 2kv, the fourth 3kv. The tenth car is 9kv

A total of 2000 V+9KV was used, not KV+2KV+3KV....

Discuss the range of x2 [0,1]g, and then correspond to the range of f, and find a

1) Define the domain x>0 f'(x)=1 x - aWhen a<0 the derivative is constant, there are no extreme points.

When a>0 x=1 a is the maximum bridge point, (1 a, lna-1) is the silver dissipation.

2) Another maximum value less than the feast is equal to -1, get:

lna-1<=-1, a>=1

It is easy to know that x=1 a is the minimum, f(x)=lnx-ax<=-1, with x=1 a

Get a>=1

The first derivative of f(x) is equal to x 2+2ax+b

Substituting the condition of knowing the potato age into this formula gives a number of traces, 1-2a+b=0, so b=2a-1

The first derivative of f(x) is (x-a) 2-(a-1) 2, which is a parabolic function and the axis of symmetry is x=-a, so it decreases monotonically in the interval (-a) and increases monotonically in the interval (-a,+ states).

f(4)=f(2)+f(2)=2

So f(x)+f(x-3) 2 can be reduced to f(x -3x) f(4) because f(x) is monotonically increasing, so x -3x 4

Solution: -1 x 4

And because f(x) defines the domain as (0,+, so x>0 and x-3>0 so when the range is 30, write f(x) as x +a 2x + a 2x by the basic inequality, and answer that when x = a 2x, f(x) obtains the minimum value.

Therefore, the increasing interval of f(x) is [ a 2 and + (a 2) 2 is required to satisfy the question

i.e. a 163(1) g(x) is the number of even functions, h(x) is an odd function, and f(x)=g(x)+h(x).

2) Any function can be expressed as the sum of an even letter and an odd function.

The definition of the odd function is f(x)=-f(-x), and since f(2)=6, f(-2)=-6, brought in, the solution is a=5.