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Personally, I think there are a few points to pay attention to when learning functions:
1。Clearly defining the domain and the value range is the premise of the correct solution function.
2。The general questions will give some basic knowledge, so it is important to understand the basic concepts clearly:
For example: odd (even) functions and their equivalent mathematical expressions (e.g., odd functions are equivalent to f(x)=-f(-x)).
Quadratic functions, power functions, exponential functions, logarithmic functions, the images and properties of these functions.
Proof of monotonically increasing (decreasing) the function over an interval.
Proof of periodic function.
3。Cultivate the thinking of combining numbers and shapes, flexibly convert the language of mathematical symbols and graphics, memorize the images and properties of basic functions, and do exercises in textbooks at the beginning.
Figuring out the above concepts, no matter how the problem is transformed, is a familiar pattern, at most plus problem-solving skills, these can be practiced through certain exercises, so learn functions to grasp the basic definition and its equivalent mathematical expression, and combine the three key factors of numbers and shapes.
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There are two types of inequalities with logarithms:
1) The base a 1, y=log(a)(x) is an increasing function: e.g. log(5)(2x+1) 2.
log(5) (2x+1) log(5) (Gausith 25).
2x+1>25。
x base 0 a 1, y=log(a)(x) is a subtraction function: e.g. log(.
log(。0<2x+1<。
1<2x<。
Solve inequalities with parameters:
When solving inequalities with parameters, attention should be paid to whether classification is necessary, such as the following situations:
When multiplying and dividing an equation with parameters at both ends of an inequality, it is necessary to discuss the positive, negative, and incomparable zeroness of the equation.
In the process of solving, when it is necessary to use the monotonicity of exponential functions and logarithmic functions, it is necessary to discuss their bases.
When solving a unary quadratic inequality containing letters, it is necessary to consider the opening direction of the corresponding quadratic function, the condition of the corresponding unary quadratic equation root (sometimes analyzed), compare the sizes of the two roots, and let the roots be (or more) but with parameters.
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Please input the y formula into cell b1:
The result is: Please enter the x formula into cell A1:
The result is:
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The first is to define the domain, which is the logarithm of log with 2 as x, then x should be greater than 0
Then the increase or decrease of the function should be judged according to the base, which is greater than 1 increase and less than 1 decrease. For example, if you give a logarithmic base of 2, the notation of inequality will not change.
1. The first question log is based on 2, the logarithm of x is 0, the log is based on 2, the logarithm of x is based on 2, and the logarithm of log is based on 2 (that is, zero), then the function is increasing, so x>1
2. If the question becomes.
log is based on 1 2 and the logarithm of x is 0
Then naturally define the domain x>0 first, and then according to the increment or decrease x
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First, the domain is defined by the properties of the logarithm.
The constant term is then converted to logarithms.
And then according to the increase or decrease of the function image, it becomes a general inequality.
For example: log2 (a+1)>1
That is: a+1>0 so a>-1
and log2 (a+1)>log2 2
Because the base number is 2>1
So: A+1>2
a>1
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4-x>0,x-1>0, 1<x<4 1,.When 0 a 1, the logarithmic function is a subtraction function, and the solution of the sum inequality is: 0 a 1, 4 x = 7 4, a 1, 1 x = 7 4
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1.The properties and images of the logarithmic function.
2.Convert to an exponential function and solve it.
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According to the analysis formula, the value that can be obtained can be obtained by directly substituting it;
According to the analytic formula, the classification discussion can solve the inequality, and finally the union of the two cases is the answer.
Solution: function, then, therefore;
function, and the inequality, so the original inequality can be reduced to , or, solution, obtainable, solution, obtainable, in summary, the set of solutions of the original inequality is .
This question examines the funny value of a function and the problem of solving inequalities in piecewise functions. For the problem of piecewise function, the idea of classification discussion and the combination of numbers and shapes is generally used to solve it, and the relevant properties are easily obtained according to the image of the piecewise function. It is a mid-range question.
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If you look at this as two functions, it is obvious that the left and right sides are even letters, and when you only see the part where x is greater than 0 touches and answers, the left function is a subtraction function, and the right function is an increase function, so either there is only one solution for the part where x is greater than 0, or there is no solution.
If we look at it, we can see that x=2 is one of the solutions, and the other is x=-2, and there are only two in total.
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I think that the question of this can be started by studying the properties of functions. The constructor is used to solve the problem of asking the positive code.
Solution: (1) Because f(x y) = f(x)-f(y)(x,y>0).
Therefore, if x=y 0, then there is f(1)=f(x)-f(x)=0i.e. f(1)=0.(2). >>>More
Since x,y is known to be positive real numbers, and x+y=1, then. >>>More
This question is a bit bluffing.
It is not difficult to solve when you note that the maximum values of [sin(x)+cos(x)-1] and [sin(2x)] are both x= 4. >>>More
1) 3-(a-5)>3a-4 (a<3)2) -5x+3<2x+1 (x>1 and 1/3)3)3-4[1-3(2-x)] greater than or equal to 59 (x less than or equal to -4)6 (1-1/3 1x) greater than or equal to 2+1/5 (10-15x) (x greater than or equal to -2). >>>More
The case of composite functions varies widely, and they are usually integrated as simple elementary functions. For example, (sinx) 2dx = 1-cos2x) 2]dx = dx 2-(1 2) cos2xdx =x 2-(sin2x 2) 2+c =x 2-sin2x 4+c can be integrated into an infinite series and then the generation will not get a simple elementary function.