# Express `f(x) = |1/(2x) - 2| - |1/(2x) + 2|` in the non-modulus form?

### 2 Answers | Add Yours

The function `f(x) = |1/(2x) - 2| - |1/(2x) + 2|`

`|x|` is equal to x if `x>=0` and it is equal to `-x` if `x < 0`

`f(x) = |1/(2x) - 2| - |1/(2x) + 2|`

If `x >= 1/4` , `f(x) = 1/(2x) - 2 - 1/(2x) - 2 = -4`

If `x <= -1/4` , `f(x) = -1/(2x) + 2 + 1/(2x) + 2 = 4`

If `-1/4 < x <1/4` , `f(x) = 2 - 1/(2x) - 1/(2x) - 2 = -1/x`

**The function `f(x) = |1/(2x) - 2| - |1/(2x) + 2|` is the same as `f(x) = [[-4, x >= 1/4],[-1/x, 1/4 > x > -1/4],[4, x<= -1/4]]` **

**Sources:**

Find separate solutions to the factors inside modulus

then **1/(2x)-2=0** => 1/(2x)=2 => **x=1/4**

and **1/(2x)+2=0** => 1/(2x)=-2 => **x=-1/4**

In number line -1/4 comes before 1/4 so **when x<-1/4** both modulus will be negative

so for **x<-1/4** = -(1/2x-2)-(-(1/2x+2)) = **4**

for **-1/4<x<1/4 **

= -(1/2x-2)-(1/2x-2) = **-1/x **

& for** x>1/4 ** = (1/2x-2)-(1/2x-2) =

**-4**

**Sources:**