Absolute value graphing (1 Viewer)

[Joshmosh2]

Hello,
"Sketch the graph of y= |x+1| - |1-x|"
From what I know, to solve is to use the ++ , +-, -+ and -- method, and somehow, the result becomes what seems to be a piecemeal function.
Any help please? thanks.

 

[Carrotsticks]

Hello,
"Sketch the graph of y= |x+1| - |1-x|"
From what I know, to solve is to use the ++ , +-, -+ and -- method, and somehow, the result becomes what seems to be a piecemeal function.
Any help please? thanks.

First, I identify what I call "fences", in this case being x=-1 and x=1. The "fences" are the critical points, in other words the values that make the inside of the absolute value equal to zero.

Now, these fences divide the number line into 3 pieces.

First piece is x>1

Second piece is -1<x<1

Third piece is x<-1

Before we start, I'll denote (x+1) by A and (1-x) by B, for simplicity when typing.

Let's consider the first piece (x>1, so for example x=2, 3, etc). When X lies in this domain, A is positive and B is negative. This means that we keep A as it is, but we FLIP B the other way around to make it positive (as the output of the absolute value must always be positive).

So for x>1, the function becomes y=(x+1)-(x-1)=2 (notice that I flipped B).

Now, let's consider the second piece (-1<x<1, so for example x=0). When X lies in this domain, A is positive and B is positive too.

So for -1<x<1, the function becomes y=(x+1)-(1-x)=2x (notice how I did not flip anything? It's because A and B were positive).

So working similarly, for piece 3, we have y=(-x-1)-(1-x)=-2.

And hence we have the piecemeal function

 

[Carrotsticks]

My apologies if this is a bit hard to follow btw, it is much better explained in person.

 

[Joshmosh2]

Thanks. That's an interesting way to look at the question. Definitely more structured than a table of values thanks!