I wasn t good at math in high school, and I wanted to know what was wrong and asked for help

You don't really digest what the teacher says, just float on the surface, and often feel that you seem to understand the lesson, but you can't do it yourself. If that's the case, I suggest you do a lot of exercises, categorize the methods of solving problems, take more notes on what the teacher says, take the time to understand more, and be sure to overcome the problems that you won't get and the wrong ones.

Mathematics, I have to say, some people are very talented. However, depending on your situation, the foundation is also quite good, and you are also very self-motivated, and you said that you don't listen very seriously in class, which is not right. What you listen to in class is different from what you study on your own, and the teacher's explanation will make you more vivid in understanding some difficult issues, so I advise you to listen well in class.

The example questions in the textbook and the exercises after class are basic things, so you must understand them. After class, you should do some difficult example problems, and learn to draw inferences from one another. Mathematics must be calculated by hand, the error rate is low, it is best to prepare an error correction book, and the problems that are easy to make mistakes are collected and solved one by one.

In short, you are only in your second year of high school, don't be discouraged, believe in yourself, there will always be a day of qualitative change.

It's not bad to score about 80 points in math in the first year of high school (out of 100 points) because the functions are more difficult. You'll need to teach yourself what you're going to learn about math during winter or summer vacation. The effect will be much better this way.

Pre-study--- lecture --- review--- homework --- review.

In fact, you can summarize more, for example, you can analyze it yourself and summarize and sort out the college entrance examination questions over the years. According to the above question types, review them by topic. Specific knowledge points such as sets and sequences. Do more questions to summarize, and the results can always come out.

Don't worry, as long as you put your heart down, you should be able to improve your grades.

After the third year of high school, there will be a second round of review, one round focusing on the foundation and the second round focusing on improvement.

Therefore, when a round of review, you must follow the teacher tightly and don't fall down, and then the basic questions left by the teacher must be able to do, and then you must do more basic questions of the same type and different methods, don't blindly pull up, that is, the foundation, lay a solid foundation, and then sort out the mistakes during this period, and then have time to do your own mistakes from scratch, if you don't have time, sort out the mistakes, and write down why you made mistakes, test the knowledge points, the aspects that need to be paid attention to, etc., without doing too much information, Because there is not much time to always look at a subject in the third year of high school, you first grasp the basics, and then try to improve in the second round, there is no need to memorize the concept clearly, but you must know what the concept is, and then it is applied to the questions, and you must do more questions. I think as long as you put your mind down and study hard, it should be no problem, and it is best to prepare a math notebook and check and fill in the gaps by yourself. Come on!

Listen carefully in class and do more questions.

Ask the question at **buy.

High school math is simple.

There should be places near high schools where you can buy books.

f（a²-1）+f（a-1)＜0

f（a²-1）＜-f（a-1)

Odd function -f(a-1)=f(1-a).

f（a²-1）＜f（1-a)

is the multiplication function a -1 1-a

Also in the defined field -11<1-a<1

The three inequalities are synapid, and the value range of a is (0,1).

It's not very difficult to get a guide, you can see if you can understand it.

f’(x)=-x^3+2x^2+2ax-2

According to the title, f(x) decreases monotonically over the interval [-1,1] and increases monotonically on [1,2], so.

f(x) has an extreme value at x=1, i.e., f'(1)= -1+2+2a-2=0, and solves a=1 2, so.

f(x)=-(1/4)x^4+(2/3)x^3+(1/2)x^2-2x-2

f’(x)= -x^3+2x^2+x-2

1. Let t=2 x, it is obvious that t>0 and know that t=2 x is an increasing function, and each x corresponds to a t, and from the title: f(2 x)=m has three different real solutions, that is, every three t of the equation f(t)=m corresponds to an m, in other words: the equation f(t)=m about t has three different real solutions at t>0.

f’(t)= -t^3+2t^2+t-2= -(t+1)(t-1)(t-2)

Let f'(t) 0 to find the increase interval of f(t), get -(t+1)(t-1)(t-2) 0, and ensure t>0, and find the increase interval of f(t) as 1 t 2

Let f'(t) 0 to find the subtraction interval of f(t), obtain -(t+1)(t-1)(t-2) 0, ensure t>0, find the image on the subtraction interval of f(t) 00 is half of the doublet shape, make the image of f(t), mark the extreme value, it can be seen that to make f(t)=m have three different real solutions, it must be -37 120, to make the image of the function y=log2[f(x)+p] have no intersection with the x-axis, only f(x)+p≠1, From the previous calculations, it can be concluded that the maximum value of f(x) is f(-1) = -5 12, i.e., f(x) -5 12

Therefore, f(x)+p p-5 12, to make f(x)+p≠1, only p-5 12<1, can satisfy the meaning of the problem, and the solution can be obtained, p<17 12

Derivative of the original function.

Just ask for a direct guide. Polynomial derivative.

f(x) is the odd function f(x) symmetry with respect to the origin of f(-x).

f(x) is an increasing function over the interval [0,2].

So the figure is 8 units loop.

25=-3*8-1 f(-25)=f(-1) in the same way f(11)=f(3) f(80)=f(0)f(-1)(-25) choose d

Since it is an odd function, it is monotonically increased on [-2,0] according to the symmetry.

f(x+8)=-f(x+4)=f(x) The period is 8f(80)=f(0).

f(-25)=f（-1）

f(11)=f(3)=-f(-1)=f(1)f(-1)so d f(-25).

Solution 1: y=1+sinx

Known: -1 sinx 1

So: 1-1 1+sinx 1+1

0≤1+sinx≤2

Therefore, the range of :y is: y [0,2].

Solution 2: y=2sin(x).

Known: -1 sinx 1

So: 2 (-1) 2sinx 2 1

2≤2sinx≤2

Therefore, the range of :y is: y [-2,2].

Solution 2: y=sin(2x)+1

Known: -1 sin(2x) 1

So: -1+1 sin(2x)+1 1+10 1+sin(2x) 2

Therefore, the range of :y is: y [0,2].

The "+1" in y=sin(2x)+1 refers to moving the image of the function y=sin(2x) up by one unit as a whole.

This is the most basic knowledge, go back and take a good look at the knowledge in the chapter on trigonometric functions.

y=1+sinx, because the value range of sinx is -1 1, so the value range of y is sinx plus 1, the value range is 0 2, y=2sinx, 2 is a multiple of sinx, that is, the value range of sinx is expanded by 2 times, which is -2 2

y=sin2x, where 2 is a multiple of the abscissa, only related to the definition domain of the function y, the value range does not need to look at it for the time being, just press sinx to take the value, the value range is -1 1

y=sin2x+1, this problem is similar to y=1+sinx, there is no need to consider the definition domain in this problem for the time being, and the value range 0 2 is directly increased

y=sin2x+1"+1", which refers to moving the y=sin2x image up by one base unit"+1"There is no specific position of the image, it just means adding 1 to the range of the function sin2x, and nothing else.

Drawing can solve everything, you just need to memorize the original basic diagram of sinx, cosx, and tanx, and take a few points (usually five points evenly) to calculate the function value for a given defined domain. Then the points can be traced and connected with smooth curves, which can be easily saved, and the value range is the value range of the y-axis. As shown in Fig

value range [-2,2].

1" is to move 1 upwards on the y-axis, and "-1" is to move 1 downwards. Actually, you can find out if you bring in the values. For example, when x=0, y=2sinx=0, and when there is "+1", x=0, y=1+2sinx=1;Changing y from 0 to 1 is moving up by 1

In addition, it is necessary to pay attention to whether the function is parity, if there is one, then only half of the drawing is sufficient, and the other half is symmetrical.

If the function is periodic, then only one period can be drawn, and the other parts will be extended to the past according to the period.

Forget to adopt).

The above value range is [0,2] [2,2] [1,1] [0,2], 1 is the whole image to translate up by one unit, the value range of sin x is [-1,1], whether it is sin 2x or sin 5x, as long as there is no other coefficient in front of sin, the value range is [-1,1], if there is a coefficient, multiply it by a few times, and add or subtract a few times later.

The value range is the maximum range in which the value of the function can be changed, and we know that the range of change of the sine and cosine functions is [-1 1], so the range of 1+sinx is [0 2]. All the parts that go to the sine have a range of -1 to 1, and as for the last question, +1 is used to translate the whole image by one unit in the positive direction of the y-axis.

Since the value range of sinx is [-1,1], the value range of y is [0,2]. The value range of 2sinx is [-2,2]; The value range of sin2x is [-1,1]; The value range of sin2x+1 is [0,2]; Plus 1 means that the entire image is panned upwards by one unit; Multiplying by 2 in front of sinx means doubling the stretch in the ordinate direction; sin2x means stretching on the abscissa. That's all, I hope you understand.

Because the range of sinx is [-1,1].So 1+sinx [-1+1,1+1], that is, the range of y is [0,2], in fact, as long as you figure out the sinusoidal function, these can be completely understood.

The value range is the distribution range of y values, and the maximum and minimum values of y can be seen when the image is drawn, and +1 means that the image as a whole is moved up by one unit.