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
Can someone please help me with piecewise functions. I know what they are but when it gets to advanced i have ttrouble.
s(x)= /x-2/ + /x-3/ + /x+1/ (/ / = 'modulus')
Write as a piecewise defined function.
Have no idea tbh.
Thanks
4 answers
'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.
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!
7 answers