If the function f x absolute value x 1 absolute value x 1 . x belongs to the set of real numbers

f(x)= |x-1| +x+1|

f(-x)=|-x-1| +x+1| = |-x+1)| x-1)| = |x-1|+|x+1| = f(x)

f(x) is an even function.

When x -1, x-1 0, x+1 0

f(x) = 1-x -x-1 = -2x When -1 x 1 is -1, x-1 0 and x+1 0

f(x)= 1-x +1+x = 2

When x -1, x-1 0, x+1 0

f(x)= x-1 +1+x = 2x

f(x) = -2x， x＜-1

2， -1≤x≤1

2x， x＞1

Method 1: If you don't have the above two questions, the easiest way is to find out "|".x-1| +x+1|(This method is used very generally).

x-1|Represents the distance from point x to point 1 on the number line.

x+1|Represents the distance from point x to point 1 on the number line.

x-1| +x+1|, which represents the sum of the distances from point x to point 1 and point -1 on the number line.

x-1| +x+1|≥2

The value range is [2,

Method 2: Now that the second question has been solved with the piecewise function, it can be written like this.

When x -1, f(x) = -2x 2

When -1 x 1, f(x) = 2

When x -1, f(x) = -2x 2

The value range is [2,

f(x)=|x-1|+|x+1|

As a slip and segmented letter, the brother of several places of the letter Lu stared at the reason.

f(x)=﹛2x,x>1

2,-1≦x≦1

2x,x

The absolute value of x-1 is the distance from x to 1.

x belongs to [-1,2], the minimum distance to 1 is 0, the maximum first sensitivity is 2 and -1, the minimum is -1, and the maximum is 1

So the value range [-1,1].

make f(x)=|x-1|+|x-2|+.x-19|The minimum value is the point where the distance between the number line is 1, 2, and the lead is empty.

This should be 10

f(10)=|10-1|+|10-2|+.10-19|

The absolute value of x x = -1

Then x=the opposite of the absolute value of x.

And because the absolute value of x is the denominator, so x ≠ 0, so x 0x can take a negative number.

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Solution: When x<-2, f(x)=-x+2)+(x-1)=-3 When -2 x<1, f(x)=x+2+(x-1)=2x+1, and -3 f(x)<3

When x 1, f(x)=(x+2)-(x-1)=3, so the range is: -3 f(x) 3

2) Please explain "inverted as.""with"Inverted at"What is the meaning?