How to find two ranges in math problems?

Is the 4+m numerator of 4 x powerless 4 or 4+m?

I'm counting by the numerator being 4.

4 x + 4 4 x + m>=2*2+m (basic inequality), i.e. 4 x + 4 4 x + m>=4+m, because the range of f(x) is r, so the true number must be able to take the positive number of the whole, that is, 4+m<0 is m<-4

log3(4 x+4 4 x+m) in the range r is the presence of m so that.

4^x+4/4^x<=-m

4 x + 4 4 x > = 2 root number (4 x * 4 x ) = 4 if and only if 4 x = 1 i.e. x = 0 is true.

Then -m>=4 i.e. m<=-4 looks.

If the value range is r, the value range of the defined domain, that is, x, is also r

To make the equation meaningful, 4 x+(4+m) (4 x)>0, i.e., 4 2x+(4+m)>0

m>-(4+4^2x)

Because 4 2>0 so 4+4 2x>4 so-(4+4 2x)<-4, so make m>-(4+4 2x) constant.

m>-4

Again, when m=-4, 4, 2x+(4+m)=4, 2x>0, so m>=-4

for r....Because after simplification, 4x points are obtained, x2+4+m is greater than 0, as long as x is not 0, then m is an arbitrary value. Take a look, right?

Can you list the function formulas first?

Because the range of two numbers is subtracted, it is necessary to judge whether it is greater than zero or less than zero, and the first range can be known to be greater than zero. At the same time, when a number takes the approximate maximum value of 3 丌 2 and a number takes the approximate minimum value 丌 2, the subtraction range of the two numbers can take the maximum value. (The maximum value of the first range is misspelled, it should be a pie.)

There is no obstacle to this question! It's the addition of a co-directional inequality.

For reference, please smile.

Because x-m 1, so -1+m, so -1+m=<1 3, and 1 2=so, 1 2=, so the range of m is [1 2,4 3].

180＜x+y＜240

180＜x－y＜-60

The two sail pants are added vertically.

0<2x<180

0 The second equation is multiplied by the state large (-1).

60 is added to the first pure judgment formula.

240<2y<420

120

Let's start with a segmented discussion of f.

Assumption a<1

When x1-a, when x=a, f=1-a

When 1>x>a f=1-x+x-a =1-a, when x>1 f=x-1+x-a =2x-1-a>1-a, so the minimum value of f is 1-a>=2 a<=-1

Assumption a<1

When x<1 f=1-x+a-x =1+a-2x>a-1 when x=1 f=a-1

When a>x>1 then f=x-1+a-x =a-1, when x>a f=x-1+x-a =2x-1-a>a-1, so the minimum value of f is a-1>=2 a>=3

In summary, a<=-1 or a >=3

The geometric meaning of this function can be expressed as the sum of the distances from the point on the number line to x=a and x=1, and a 3 or a -1 is required for the function value to be greater than or equal to 2

First of all, let a = (1-b)*k + b = b(1-k)+k and the range of a is the range of a.

So it can be solved:

1]0[2]-1[1] Yes. 0[2] Yes. b-1k∈(-1,0)；

b∈(0,; 2b-1∈(-1,0)

So the maximum range of a is:】 1, i.e., -1<(1-b)k+b<

The conditions for the critical point are:

When k=-1, b=0, a=-1

When k=0, b=, a=

x1x2＞1/2

Is it a product, or a separation