Ask a question about finding the range of a function, a question about finding the domain of a funct

This question may be better if it is used in a discriminative method.

Remove the denominator and sort out (y-3)x 2+(y-3)x-(y+1)=0, the above equation has a solution to the equation about x, so the discriminant formula =(y-3) 2+4(y-3)(y+1)>=0 , and y-3 ≠ 0 , decompose (y-3)(5y+1)>=0 , and solve y<= -1 5 or y>3.

You can use the separation constant method, but discuss the range of values for the denominator.

Let t=x 2+x-1=(x+1 2) 2-5 4>=-5 4y=(3x 2+3x-3+4) (x 2+x-1)=(3t+4) t=3+4 t

Because t>=-5 4, so 1 t>0, or 1 t<-4 5 so y>3 or y<3-16 5=-1 5

Solution: The answer to the two-digit solution upstairs is wrong, and the method of separating constants cannot be used here. Because in fractions, the denominator.

x 2+x-1=0 has two different real roots, i.e. x1=[-1+5 (1 2)] 2 x2=[-1-5 (1 2)] 2

For molecules 3x 2 + 3x + 1 = 3x 2 + 3x-3 + 4 = 3 (x 2 + x-1) + 4

The original function y=(3x 2+3x+1) (x 2+x-1)=3+4 (x 2+x-1).

Let u=x 2+x-1 when x tends to [-1+5 (1 2)] 2, u tends to 0 y tends to +infinity.

When x tends to [-1-5 (1 2)] 2, u tends to 0 and y tends to -infinity.

Therefore, the range of the function is: (negative infinity, positive infinity).

The domain is defined as r

AX 2-2X+A 0 was established.

a 0 and =4-4a 2 0

So the range of a 1 is r

ax^2-2x+a

I can take all the numbers greater than 0.

a 0 and =4-4a 2 0

So 0 a 1

The definition of the BAI domain is R

ax 2-2x+a 0 is always established, that is, only a 0 and dao=4-4a 2 0 are satisfied

You can get a 1 specifically

The range is rax 2-2x+a

You can take all positive numbers.

So only a 0 and =4-4a 2 0 are needed

Solution: 0 a 1

1、y=4^x+2^(x-1)+1

2、y=3^x / 3^x +1)

Solution: 1, y=4 x+2 (x-1)+1

2^x)^2+(2^x)/2+1

Let t=2 x t>0

Then y=t 2+t 2+1

t+1/4)^2+15/16

Because t>0

So the minimum value of y is 1 but you can't get it.

So the value range is (1, positive infinite reputation and suspicion).

2、y=3^x / 3^x +1)

Let t=3 x t>0

then y=t (t+1).

1/(1+1/t)

Because t>0 so 1 t> shed number 0

1+1/t>1

So y<1

So the range is (0,1).

y=√[(x+3)²+0+4)²]x-5)²+0-2)]²

So y is the sum of the distances from a point p(x,0) to a(-3,-4),b(5,2) on the x-axis.

Apparently when the PAB is collinear and the P is between ABs, it is minimal.

Here ab is on both sides of the x-axis.

So it's eligible.

The distance and minimum are |ab|= (8 +6 = 10, so the minimum value is 10

y= (x 2+6x+25)+ x 2-10x+29)= [(x+3) 2+4 2]+ x-5) 2+2 2] Therefore, y is the sum of the straight-line distances between (x,0), (3,4) and (x,0), (5,2).

The linear equation for (-3,4),(5,2) is: y=kx+b, then: -3=4k+b, 5=2k+b

k=-4,b=13/4

y=-4x+13

4x+13=0

x=13/4

The straight-line distance between two points is the shortest.

The minimum value of y is:

y=√[(x+3)^2+4^2]+√x-5)^2+2^2]=√[(13/4+3)^2+4^2]+√13/4-5)^2+2^2]=1/4(√421+√113)

To give an idea, x 2+6x+25+ x 2-10x+29 can be broken down into x 2+6x+9+16 + x 2-10x+25+4 = x (x+3) + 16 + x-25+4 + x+4

There are a lot of my rolls, but to type out, it's very troublesome, you call the upstairs with the Sogou input method, there are symbols.

There are a lot of math symbols in this question that I won't type out with a keyboard

There is no fixed method or model. However, the commonly used methods are: (1) Direct method:

Starting from the range of variable x, the value range of y=f(x) is deduced. (2) Matching method: The matching method is the basic method to find the value range of the "quadratic function class", such as the value range of the function of f(x)=af (x)+bf(x)+c, you can use the matching method (3) Inverse function method: The inverse relationship between the definition domain and the value range of the function and its inverse function is used to obtain the value range of the original function through the definition domain of the inverse function.

Functions of the form y=cx+d ax+b(a≠0) can be used as an inverse function. In addition, this type of function range can also be solved using the Separation Constant Method. (4) Commutation method:

Using algebra or trigonometric substitution, the given function is transformed into another function whose range is easily determined, so as to obtain the value range of the original function. Functions of the form y=ax+b root number cx+d (a, b, c, d are constants, and a≠0) are commonly solved in this way. Let's give some examples!

1) y=4-root number 3+2x-x This problem has to use the matching method: from 3+2x-x 0, we get -1 x 3y=4-root-1(x-1) +4, when x=1, ymin=4-2=2

When x = -1 or 3, ymax = 4The function range is [2,4] (2)y=2x+root 1-2x This problem uses the commutation method: Let t=root 1-2x(t 0), then x=1-t 2 y=-t +t+1=-(t-1 2) +5 4, when t=1 2 i.e., x=3 8, ymax=5 4, there is no minimum value.

The range of functions is (- 5 4) (3)y=1-x 2x+5 y=-1 2+7 2 2x+5, 7 2 2x+5≠0, y≠-1 2