Why is the function Y X 2 1 called a function from a set of real numbers to a set of real numbers

This statement is incorrect.

y=x 2+1 can be thought of as a function of the set of real numbers to the set of real numbers, as well as a function of the set of integers to the set of integers, as a function of the set of complex numbers to the set of complex numbers, or even as a function of the set of quaternion numbers to the set of quaternions.

First of all, you need to understand the definition of the concept of function. In high school math textbooks in our country (at least that's the case with the Renjiao Society), the definition of a function is:

For two non-empty sets A and B, if there is a unique element B corresponding to any element A in set A according to a certain correspondence rule f, then F is said to be a mapping from set A to set B, denoted as F: A B.

When a and b are both sets of numbers, mapping f is also called a function of set a to set b.

There are three elements of a function, the definition domain, the value range, and the corresponding law, all of which are indispensable.

It would be inappropriate to give only a correspondence without specifying the definition and value ranges.

In college, if you take courses such as Fundamentals of Mathematics or Mathematical Logic or Discrete Mathematics or Algebraic Structures, you will be more strictly defined as a law (relation).

Odd power functions, exponential functions, and trigonometric functions can all be thought of as functions from the set of real numbers to the set of real numbers.

Because the definition domain of this function is a real number, and the corresponding range of values that can be obtained is also a real number, it is called a function from the set of real numbers to the set of real numbers.

Others, e.g. y=x, y=sinx, y=lnx, y=|x|And so on.

But things like y=[x] ([is a Gaussian mark, [x] is the largest integer not greater than x), y=sgnx (symbolic function, the value of the function y takes the positive or negative sign of x or 0), etc.

Solution: x +xy + y = 1

1=x²+y²+xy≥2|xy|+xy

xy 0, then 1 3xy, get xy 1 3xy<0, then 1 -xy, get xy -1 so -1 xy 1 3

x²-xy+y²

1-xy-xy

1-2xy1≤xy≤1/3

2/3≤-2xy≤2

Therefore, the value range of x2-xy+y2 is [1 3, 3].

Since x2+xy+y2=1

Therefore, xy=1-(x2+y2) is substituted for x2-xy+y2, which is to find the value range of x2-[1-(x2+y2)]+y2 and simplify x2-[1-(x2+y2)]+y2.

2(x2+y2)-1

That is, find the range of 2(x2+y2)-1.

Since x2+y2>=0

So 2(x2+y2)-1>=-1

Therefore 2(x2+y2)-1>=-1

Hello! Obviously, x=-3a<0 is outside the defined domain and should be discarded.

So x=a can be h(x)=f(x)-g(x).

h(x)'=x+2a-3a^2/x

Observations show that when x=a, h(x).'=0

and 0a,, h(x).'>0

The minimum value is h(a)=f(a)-g(a).1）

Substitute x=a with :

f(a)=g(a)

That is, f(a)-g(a)=0 is substituted into (1) to obtain:

The minimum value h(x)=0

Therefore f(x)>=g(x).