gomes. said:
Im stuck in the 4 red boxes I've highlighted.
1st box: what is the method/quickest way to draw the graph?
You could plot points. That's not what I would do, though.
I would draw a graph of y = |x - 1| (has a vee shape, with the low point at (1, 0)). Then I would draw a graph of y = |x - 1| - 2, which is a translation downward of the previous graph. Finally, I would draw a graph of y = ||x - 1| - 2|. Any parts of the previous graph that are below the x-axis should be reflected across the x-axis. Any parts of the previous graph that are above the x-axis should be left alone.
gomes. said:
2nd box: what does it mean by open interval?
An open interval is an interval that does not include the endpoints. There are two ways to denote intervals: by inequalities such as 2 < x < 4 and 0 <= x <= 5; by interval notation such as (2, 4) and [0, 5]. The first inequality and its equivalent interval notation represent an open interval. The second inequality and its equivalent interval notation represent a closed interval. An interval can also be half-open and half-closed if one endpoint is included and the other not included in the interval.
gomes. said:
3rd box: how would i sketch the preimage? what does preimage really mean?
The preimage of a function is the set of all numbers or points that are valid inputs for the function. What they seem to be asking for in this problem is the set of all points in the plane such that f(x, y) = 4. Note that for this function there are a lot of points that satisfy the equation f(x, y) = 4.
gomes. said:
4th box: how would i prove it for open interval?
Before tackling how you would prove it, what do you think |A| means in this problem?
gomes. said:
[PLAIN]http://img189.imageshack.us/img189/526/untitlex312x31d.jpg[/QUOTE]