Question

Hi, Is it possible (and if it is, how) to find the equation of a function with absolute values when I know a graph of that function?
I mean functions like f(x) = |x-3| + |2-5x|

Answer 1

....so you're looking to find the equation of a function......given an equation of a function....?

Answer 2

I only know a graph.

Answer 3

..but you gave an equation....

Answer 4

f(x) = g(x) + h(x)

Answer 5

^he/she probably made more sense than i did

Answer 6

lol, im a he/she ??

Answer 7

sorry...i forgot....."it"

i just always get the feeling you're very close to human-ness so i call you a he/she

Answer 8

us similated answering programs have rights too ;)

Answer 9

this would be easier if we had a graph to go by

Answer 10

In particular how to find the equation of

Answer 11

I know it can be written as a picewise function , but it is possible to write it as equation with absolute values.

Answer 12

the vertexes might be able to give us a clue as to that

Answer 13

-1 = g(1) + h(1)

1 = g(0) + h(0)

5 = g(2.5) + h(2.5)

5 = g(3) + h(3)

Answer 14

OK but the equation might contain more than two absolute values in it.

Answer 15

yeah, i was thinking of that as well; but this is just a general idea

Answer 16

im wondering if we can assume that the given slopes are the averages parts

Answer 17

also the absolute values need not be linear either

Answer 18

but it prolly helps if they are linear

http://www.wolframalpha.com/input/?i=y%3D+%7Cx%5E2-6%7C

Answer 19

Equation might be in the form:
\[
f(x) = a_1x + b_1 + a_2|x + b_2| + a_3|x + b_3| + \ldots
\]

Answer 20

where \[a_1,a_2, a_3, \ldots, b_1, b_2, b_3, \ldots
\] can be any real numbers

Answer 21

im curious if we can conform this by using a method of zeroing out coeffs

Answer 22

\[y = c_1~x|x-1||x-2.5|+c_2~x|x-1||x-3|+c_3~x|x-2.5||x-3|\]\[\hspace{3em}+c_4~|x-1||x-2.5||x-3|\]

Answer 23

it doesnt quite fit, but its got a similar shape :)

http://www.wolframalpha.com/input/?i=y%3D5x%7Cx%E2%88%921%7C%7Cx%E2%88%925%2F2%7C%2F3%2B8x%7Cx%E2%88%921%7C%7Cx%E2%88%923%7C%2F3-x%7Cx%E2%88%925%2F2%7C%7Cx%E2%88%923%7C%2F3+%2B2%7Cx%E2%88%921%7C%7Cx%E2%88%925%2F2%7C%7Cx%E2%88%923%7C%2F25

Answer 24

\[f(x)\quad \begin{matrix}-2x+1~;~x\in(-\infty,1)\\4x-5~;~x\in \left[1,\frac52 \right)\\5~;~x\in \left[\frac52,3 \right] \\2x-1~;~x\in (3,\infty)\end{matrix}\]
i wonder, since there are 4 piecewise parts, if this could be comprised of 4 absolutes value function

Answer 25

I think, that it can be written as a sum of 3 functions of a form a|x + b|

Answer 26

do you have material that goes with this?

Answer 27

if its just some idea that you are working with, thats fine too; but if there is material associated with it, that could help to focus the thoughts a little :)

Answer 28

No, I don't. I'm just learning about absolute value functions and i do some exercises in a book for graphing. I was curious if i can reverse the "process".

Answer 29

im thinking that since we have 4 different slopes on the graph, that we would need to average 4 different points in each interval to match. just an idea.

what is your notion for it being 3?

Answer 30

Just messing around with GeoGebra.
When it has 3 absolute values it has 3 breaks.

Answer 31

3 breaks yes, but i cant see a way for one of those breaks to be horizontal

Answer 32

y = |x+1| + |x-1| gives us a horizontal; anytime we provide a shift, we can get a horizontal section

Answer 33

It looks that there is a horizontal part in some interval if sum of two absolute value functions have opposite slope in that interval.

Answer 34

Or sum of their slopes is zero.

Answer 35

In general slope of the function in some interval is equal to sum of slopes of functions in that interval.

Answer 36

http://www.wolframalpha.com/input/?i=y%3D2%282%7Cx-1%7C-1+%2B+4%7Cx-5%2F2%7C%2B5+-+2%7Cx-3%7C%2B5%29-19

that looks promising if you can work out the dynamics between them

Answer 37

http://www.wolframalpha.com/input/?i=y%3D3%7Cx-1%7C-1+-2%7Cx-5%2F2%7C%2B3%2B%7Cx-3%7C-2

i cant tell, but this is sooo close

Answer 38

that was it :)

http://www.wolframalpha.com/input/?i=derivative+3%7Cx-1%7C-1+-2%7Cx-5%2F2%7C%2B3%2B%7Cx-3%7C-2

Answer 39

i wish i knew what i did, but i kept working with the vertices somehow

Answer 40

I made some progress.
Assume that the function consists of sum of 3 functions of a form a|x + b|
So it will be
\[ f(x) = a|x + b| + c|x + d| + d|x+e| \]
where a,b,c,d,e are real numbers.
Function has breaks at points 1, 2.5 and 3, hence
\[ f(x) = a|x - 1| + c|x - 2.5| + d|x-3| \]
And then I can divide the function into 4 intervals
Let them denote
\[
I_1 = (-\infty, 1) \\ I_2 = [1, 2.5) \\ I_3 = [2.5, 3) \\ I_4 = [3, +\infty)
\]
As I discovered before, slope of the function is the sum of slopes of function the function is build from.
So I can solve for slope in each interval. Thaen I get system of equations
\[ a + c + e = 2 \\
a- c - e = 4 \\
a + c - e = 0 \]
From which I get
\[ a = 3 \\
c = -2 \\
e = 1\]

And therefore the function is
\[\Large f(x) = 3|x-1| - 2|x-2.5| + |x-3| \]
http://www.wolframalpha.com/input/?i=+f%28x%29+%3D+3 |x-1|+-+2|x-2.5|++%2B+|x-3|

Answer 41

that seems more systematic than the approach i took -- some skillm and alot of luck ;)

Answer 42

how did you arrive at this setup ??

a+c+e=2
a−c−e=4
a+c−e=0

Answer 43

also you have a 4th slope, how did you figure which one to ignore?

Answer 44

the fourth one was -(a+c+e) = -2 which is -1 multiple of the first

Answer 45

Ok in the first interval the equation is
\[ -a|x - 1| - c|x- 2.5| - e|x - 3| =
-(a + c + e)x + a + 2.5c + 3e\]
but i only need a slope which is
\[ -(a+c+e) \]
from the graph you can see that slope in that interval is -2
hence the first equation is
\[ -(a+c+e) = -2 \]
Slopes in other intervals are done in the same way

Answer 46

Oh, in the previous post the equation should be without absolute values
\[
−a(x−1)−c(x−2.5)−e(x−3)=−(a+c+e)x+a+2.5c+3e
\]

Answer 47

a|x-1| +b|x-2.5| +c|x-3| = -2
a|x-1| +b|x-2.5| +c|x-3| = 4
a|x-1| +b|x-2.5| +c|x-3| = 0
a|x-1| +b|x-2.5| +c|x-3| = 2

i still cant see how you get from that to the setup

a+c+e=2
a−c−e=4
a+c−e=0

Answer 48

im glad you did of course :) but i cant see a reasoning to it

Answer 49

im using abc for your ace of course

Answer 50

In the first interval expression in all three absolute values is negative
so you can write the equation as
\[ −a(x−1)−c(x−2.5)−e(x−3) \]
slope is -(a+c+e)
in the second interval is x- 1 non negative and x - 2.5 and x- 3 are negative so it can be written
as \[a(x-1) - c(x-2.5) - e(x-3) \]
slope is a-c-e
In the third interval x-1 and x-2.5 are non-negative and x-3 is negative therefore it can be written as
\[ a(x−1)+c(x−2.5)−e(x−3) \]
slope is a+c-e
Last, in the fourth equation are all expression in absolute value non-negative, and can be written as
\[ a(x−1)+c(x−2.5)+e(x−3) \]
slope is a+c+e

From which i get the system of equations
\[
-(a+c+e) = -2 \\
a-c-e = 4 \\
a+c-e = 0 \\
a+c+e = 2
\]

Answer 51

i would conjecture from that, that we start out assuming that the slopes are all equal, then switch them out one by one till, in the different intervals we have the slope of one of them flipping ....

Answer 52

equal meaning of course in direction + or -

Answer 53

spose we had 5 slopes, such that the ends are "equal" but opposite

then:

a+b+c+d = s1
a+b+c-d = s2
a+b-c-d = s3
a-b-c-d = s4

is that a fair idea?

Answer 54

i wonder if we can construct a summation of these things such that the ends have different slopes ....

Answer 55

The method of division into intervals when you solve an equation and solve in each interval is something I learn in school and I've never seen in English book or website.
Is that method taught in English-speaking countries?
If not I can try to explain it in more detail.

Answer 56

im not sure what is in the schools these days; ive been out of class for 25 years or so :)

Answer 57

if the ends have different slopes; say -2 on the left and 3 on the right, then you cant construct it with absolutes. If both ends have the same slope that differ by a sign; then it appears you can

Answer 58

if you use an extra third vertex at say (inf,inf) maybe :)

Answer 59

Then it probably has
\[ a_1 \neq 0, \\
b_1 \neq 0 \]
in the equation
[\ f(x)=a_1x+b_1+a_2|x+b_2|+a_3|x+b_3|+\ldots \]

Answer 60

your first delimiter is flipped about, spose to be \[

Answer 61

\[ f(x)=a_1x+b_1+a_2|x+b_2|+a_3|x+b_3|+\ldots \]

Answer 62

ah yes, that plain old linear in the front does help to get different slopes

Answer 63

as long as a1 not= 0, i can form different slopes; the b1 is superfluous since that only adjusts the height overall and not the slope