Can someone please help me with piecewise functions. I know what they are but when it gets to advanced i have ttrouble.

not sure what you mean by "gets to advanced."

Not sure why you used slashes ? instead of the real symbol |.

so, we have

s(x) = |x-2| + |x-3| + |x+1|

Since you asked no questions about it, nor provided the division of the domain among the parts, I'll just present the graph for you to ponder:

http://www.wolframalpha.com/input/?i=%7Cx-2%7C+%2B+%7Cx-3%7C+%2B+%7Cx%2B1%7C+for+-3+%3C%3D+x+%3C%3D+5

'Write as a piecewise defined function.'

As you can see from Steve's graph, your "critical" values are

x = -1,2,3
so you have the following line segments
1. for x ≤ -1 :
let x = -2, y = 10
let x = -1 , y = 7 , slope = -3
y-7 = -3(x+1)
-----> y = -3x + 4

2. for -1 ≤ x ≤ 2
let x = -1 , y = 7
let x = 2 , y = 4 , slope = -1
y - 4 = -1(x-2)
y = -x + 6

3. for 2 ≤ x ≤ 3
......

4. for x ≥ 3
.....

You do the remaining two relations.

then
s(x)
= -3x + 4 , for x < -1
= -x + 6 , for -1 ≤ x < 2
= ......... , for 2 ≤ x ≤ 3
= ......... , for x > 3

notice that I used the common points or linkage points such as (-1,7) in only one of the relations, but not in both
The choice of which one is up to you.

Good job. I should have seen that the vertices would be where to divide the lines. My bad!

To write the given function as a piecewise defined function, you need to break it down into multiple cases based on the values of x.

For the given function s(x) = |x-2| + |x-3| + |x+1|, we can break it down into three cases based on the signs of the expressions inside the modulus:

Case 1: When x <= -1
In this case, x is less than or equal to -1. For these values of x, the expressions inside the modulus simplify to negative values. Therefore, we can write the function as:
s(x) = -(x-2) - (x-3) - (x+1)

Case 2: When -1 < x <= 2
In this case, x is between -1 and 2. For these values, the first and second expressions inside the modulus simplify to their positive values, while the third expression simplifies to a negative value. Therefore, we can write the function as:
s(x) = (x-2) + (x-3) - (x+1)

Case 3: When x > 2
In this case, x is greater than 2. For these values, all three expressions inside the modulus simplify to their positive values. Therefore, we can write the function as:
s(x) = (x-2) + (x-3) + (x+1)

Combining all these cases, we can write the piecewise defined function as:

s(x) =
-(x-2) - (x-3) - (x+1) if x <= -1
(x-2) + (x-3) - (x+1) if -1 < x <= 2
(x-2) + (x-3) + (x+1) if x > 2

I hope this helps!