How do you get a high one piece function, can you explain to me his definition???????

Definition. piecewise functions; There are different correspondence rules for different ranges of values of the independent variable x, and such functions are usually called piecewise functions. It is a function, not a few functions: a defined domain of a piecewise function.

is the union of the domain defined by each segment function, and the value range is also the union of the domain of each segment function.

Type. 1. The mathematical expression around the demarcation point is the same, but the value of the function at the demarcation point is defined separately (Example 1).

2. The mathematical expressions around the cut-off point are different (Example 2).

Example. Example 1 a shopping mall holds a prize shopping activity, every 100 yuan of goods to get a lottery ticket, every 1000 lottery tickets for a group, numbered 1 to 1000, of which there is only one winning grand prize, the grand prize amount of 5000 yuan, when the lottery, the winning grand prize number is 328, then, a lottery ticket to get the grand prize Y yuan and the number X number of the functional relationship is expressed as.

0 ,x≠328

y={ 5000, x=328

Example 2 A store sells watermelons, if the weight of a watermelon is less than 4kg, then the sales ** is yuan kg; If it is 4kg or more, then the sales ** is yuan kg, then, the functional relationship between the sales revenue of a watermelon y yuan and the weight xkg is expressed as.

0〈x〈4y={ ,x≥4

Piecewise function question type.

Since the textbook does not explicitly give the piecewise function.

The definition of problem only appears in the form of example problems, and many students have a superficial and vague understanding of it, so that students often make mistakes in solving problems. This paragraph introduces several types of problem types of piecewise functions and their solutions for your reference.

Pick D. The piecewise function is actually within a segment of the value of x, y or f(x) will have different expression values, which is reflected in the different functions on your test paper. The landlord only needs to bring the expression of f(x) into f(x)=3 to see if the value calculated by x is within the corresponding interval.

Bring each value in to solve, and then find the value of x to see if it is within the range of its constraints, if not, it will be discarded, if it is, x is its value, and each function will be solved again.

Senior 1 Mathematics: The value of the parameters of the piecewise function also needs to be discussed in sections.

f(x)=x 2 (-1 x 1), then 0<=f(x)<=1

f(x)=2 (x 1 or x 1) So, the range of f(x) is [0,1]u

If the equation f(x)=a-1 has a solution, then there is:

0<=a-1<=1 or a-1=2

The solution yields 1<=a<=2, or a=3

When encountering such a piecewise function evaluation, it is calculated layer by layer.

If you don't understand anything about this question, please feel free to ask.

Good luck with your studies!

Solution: f(x)=x+1 (4x) 2 [x*1 (4x)]=1,(x>0, if and only if x=1 (4x) i.e. x=1 2, take "="), i.e., when x>0, f(x) 1, and.

When a>1, f(x)-a=0 has 2 real roots on (0,+), when a=1, f(x)-a=0 has 1 real root on (0,+, and when a<1, f(x)-a=0 has no real roots on (0,+;

f(x)=-x -4x-1(x 0), is the part of the parabola x 0 with the opening pointing downward and the axis of symmetry is x=-2, when x=-2, f(x)=3, when x=0 and x=-4, f(x)=-1, and.

When a=3, f(x)-a=0 has 2 heavy real roots on (- 0], when -1 a<3, f(x)-a=0 has 2 different real roots on (- 0), and when a<-1, f(x)-a=0 has only 1 real root on (- 0);

To sum up, the sufficient and necessary conditions for the four real roots of the equation f(x)-a=0 are that the equation f(x)-a=0 has 2 real roots on (- 0), and there are 2 real roots on (0,+, and the value range of the real number a satisfying the condition is that f(x) is the right branch of the double hook function when 10, and f(x) is part of the parabola when x 0.

<> the conditions for the destruction of the source of the question. Don't dig up the surplus, don't just look at the formula.

It was quickly unraveled with substitution. As follows:

Because: f(a)+f(1)=f(a)+2 1=0, so f(a)= -2 and because for any real number a, f(a)=2 a>0; So f(a)=a+1=-2

Solve this equation to obtain: a= -3

f(x) is the piecewise function:

f(x)=x+1,-1≤x<0;

f(x)=-x,0≤x≤1.

Define the domain. Draw according to the value of the defined field.

For example, if the function is divided into x, when the function is less than 5, the function is y=x+4, and when the function is greater than or equal to 5, the function is y=x-6

I can't see clearly, I really don't know.