Hi
For y = abs(x+2) + abs(x) + abs(x-2), what are the cases?
I know if there were only two abs, there would be three cases to work out but I have no clue how many there will be for this question and what they might be.
Thanx
Dear xwrathbringerx,
the abs(x) function behaves "irregular" at x=0 that is x=0 is a singular point.
Therefore the function abs(x-2) has sigularity at x=2, hasn't it?
So in your problem you have to divide into 4 cases because there are 3 singular point.
If $\displaystyle x\le -2$ then $\displaystyle y=-(x+2)-x-(x-2)=-3x$.
If $\displaystyle -2\le x\le 0$ then $\displaystyle y=(x+2)-x-(x-2)=-x+4$.
If $\displaystyle 0\le x\le 2$ then $\displaystyle y=(x+2)+x-(x-2)=x+4$.
If $\displaystyle 2\le x$ then $\displaystyle y=(x+2)+x+(x-2)=3x$.
The graph is attached.
Hello, xwrathbringerx!
Another approach . . .
Make a sketch . . .For $\displaystyle y \:=\:|x+2| + |x| + |x-2|$, what are the cases?
The graph of .$\displaystyle y \,=\,|x|$ .is a $\displaystyle \vee$ with its vertex at the origin.
The graph of .$\displaystyle y \,=\,|x-2|$ .is the same graph moved 2 units to the right.
The graph of .$\displaystyle y \,=\,|x+2|$ .is the same graph moved 2 units to the left.
Code:\ \ | / / \ \ | / / \ \ | / / \ \ | / / \ \|/ / \ \ * / / \ \ /|\ / / \ \/ | \/ / \ /\ | /\ / \ / \|/ \ / ---------*----*----*------- -2 | 2
Then add the ordinates.
We get the graph that james_bond so thoughtfully provided.