find the domain,range and graph the following function:
f(x)=|x+1|/x-1

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Clarification of Question by ryukyu90-ga on 02 May 2005 06:24 PDT

I need to find the piecewise functions, so that i can plot the points
and get the graph. Note that I already know how it should look, but i
need to learn how to get to that point. Also , lots of teachers have
rewritten the f(x)=|x+1|/x-1, as following:

-1+|1+x|/x, how is it possible to arrive @ this equation form the original function?

If you've got a TI, you can use it to graph it pretty easily...  Hit
the "Y=" key (top left corner) and in the first blank enter
"abs(x+1)/x-1".  The "abs()" function can be found by pressing the
"Math" key (under Alpha), then the right arrow, and then enter.  As
far as domain and range, well, the graph should tell you all you need
to know.

If you don't have a TI, get one.  XD

If you mean 
    f(x)=|x+1|/(x-1)
as opposed to 
    f(x)=(|x+1|/x) - 1 
then the domain is all real numbers from negative infinity to positive
infinity except the number 1, because at x=1 the function is undefined
(divide by zero error). The range is from negative infinite up to zero
inclusive, plus from 1 (not inclusive) up to positive infinity. A
graph should make it clear why. Kind of a neat-looking function with
the hitch at x=-1.

To answer you last question first, let's look at the actual equation.
y=(|x+1|/x)-1.  The important thing to remember here is that addition
is commutative.  So we can take the |x+1|/x and the -1 and switch
them, getting y=-1+|x+1|/x.  Then we do the same thing inside the
absolute value signs, again remembering that addition is commutative
and so that |x+1|=|1+x|.  This gives us the final equation of
y=-1+|1+x|/x.

You want to look at an absolute value as a sort of piecewise-defined
thing in and of itself.  Let's take this bit by bit.

If we had the graph y=|x|, it would look like that v with which you're
probably very familliar by now.  What we're actually doing is saying,
when the value inside the absolute value sign is already positive
(looking at our domain), the output is like y=x (looking at our
function/range).  But when the value is negative (new piece, new
domain), the absolute value switches the sign, making it positive, so
the graph actually looks like y=-x over that domain (keeping in mind
that the negative of a negative is a positive).  So from negative
infinity to 0 exclusive, y=-x, and from 0 exclusive to infinity, y=x. 
At 0, the lines intersect, so you can pick one of the two to be
inclusive at 0 (but not both or it's not a function), but it doesn't
matter which.  This is our piecewise-defined function.

Now, what about y=|x+1|?  Let's do nearly exactly the same thing. 
When is the input positive?  So long as x+1>0, or x>-1.  So for x
between -1 exclusive and infinity, our function looks like the line
y=x+1.  The input is negative when x+1<0, or x<-1, in which case we
awnt to do the same thing as we did in the last case, namely, to
switch the sign ourselves.  If (x+1) is negative, -(x+1) is positive. 
This may be easier to write as -x-1.  So from negative infinity to -1
exclusive, y=-x-1.  I think you can probably put together the function
from this point.

y=|x+1|/x has the same general bounds as the previous function, but we
reach a problem at 0 as well, because you cannot divide by 0.  Other
than that, though, it's exactly the same.  The inner value is still
already positive over x>-1 (so long as we're not at 0 where we run
into division problems), giving the function y=(x+1)/x for x between
-1 and 0 exclusive and 0 exclusive and infinity.  And likewise, over
the domain x<-1, the absolute value is negative, so the function is
y=(-x-1)/x.

Subtracting a 1 from there does nothing to the domain, so you just
need to adjust your functon accordingly.  But I need to go to school
now, so I'll stop here.  Please post if you'd like further
clarification... I'd hate to keep babbling if you already understand
about what I'm talking.

Sorry for my broken English! I'll try to study your function.

"[-" = "to belong to".
"(/)" = "an empty set".
"=/=" = "unequal".
"R" = "the set of rational numbers".
"D(y)" = "the domain of the function y(x)".
"E(y)" = "the range of the function y(x)".
"y'(x)" = "the derivative of the function y(x) with respect to x".
"y''(x)" = "the second derivative of the function y(x) with respect to x".

y(x)=-1+|1+x|/x.
         _
        |  1, x>0
sign(x)={ -1, x<0
        |_ 0, x=0.

|x|=sign(x)*x.
|x|'(x) = sign(x), x =/= 0.

|x+1|=|1+x|.
|x+1|/x-1 = -1+|1+x|/x.

D(y)=R\{0}.

lim             y(x) = y(a).
x -> a =/= 0, +0, -0

sign(1+x)=s.

(u(x)/v(x))'(x)=(u'(x)*v(x)-v'(x)*u(x))/v(x)^2.
(x^n)'(x) = n*x^(n-1).

y'(x) = (s*x-|1+x|)/x^2 = (s*x-s*(1+x))/x^2 = -s/x^2 = -s*x^(-2), x =/= -1.
y''(x) = 2*s*x^(-3), x =/= -1.

y(x)=0 <=> -1+|1+x|/x = 0 <=> (|1+x|=x, x =/= 0) <=> x [- (/).
(y'(x)=0, x =/= -1) <=> (x=-1, x =/= -1) <=> x [- (/).
(y''(x)=0, x =/= -1) <=> (x=-1, x =/= -1) <=> x [- (/).

I1 = (-infinity; 0).
I2 = (0; +infinity).
D(f) = I1 U I2.

I1:
************
y(x)<0.

lim y(x) = -infinity.
x -> -0

************

I2:
************
y(x)>0.

lim y(x) = +infinity.
x -> +0

************

J1 = (-infinity; -1).
J2 = (-1; 0).
J3 = (0; +infinity).
D(f) = J1 U {-1} U J2 U J3.

J1:
************
y'(x)>0 => y(x) increases.
y''(x)>0 => y(x) is convex downwards.
************

{-1}:
************
y(-1)=-1.
************

J2:
************
y'(x)<0 => y(x) decreases.
y''(x)<0 => y(x) is convex upwards.
************

J3:
************
y'(x)<0 => y(x) decreases.
y''(x)> => y(x) is convex downwards.
************

E(y)=(-infinity; -1] U (0; +infinity).

I hope that i don't make any mistake and help you...
You can see the graph of your function here (4,37K):

http://www5.webfile.ru/get/1192669759/yx.GIF

Thanx for the interesting problem!