Expert: Paul Klarreich - 12/4/2009

Question
Paul,

We are working on a packet for math and have a couple questions we can't figure out. Here they are:

Let f be the function given by f(x)=ln(abs(x/x-1)). We found the domain of (-inf, 1)U(1, inf) but can't figure out part two: Find the value of the derivative of f at x=-1. How does the absolute value affect the derivative? Part 3 is to write an expression for f^-1(x), where f^-1 denotes the inverse function of f. Can you find an inverse function of an absolute value function?

Also, how do you find the range of f(x)=((abs(x))-2)/(x-2)? We know it is -1<x< or equal to 1, but we're not sure how to prove it. We said that f(x)=1 for all x> or equal to 0, but what do you do about x<0?

Thanks so much.

--Sam, Meghan, Amy Rose

Answer
Questioner: Sam, Meghan, and Amy Rose
Country: United States
Category: Calculus
Private: No
Subject: AP Calculus BC
Question: Paul,

We are working on a packet for math and have a couple questions we can't figure out. Here they are:

Let f be the function given by f(x)=ln(abs(x/x-1)). We found the domain of (-inf, 1)U(1, inf) but can't figure out part two: Find the value of the derivative of f at x=-1. How does the absolute value affect the derivative? Part 3 is to write an expression for f^-1(x), where f^-1 denotes the inverse function of f. Can you find an inverse function of an absolute value function?

Also, how do you find the range of f(x)=((abs(x))-2)/(x-2)? We know it is -1 < x <= 1, but we're not sure how to prove it. We said that f(x)=1 for all x> or equal to 0, but what do you do about x<0?

Thanks so much.

--Sam, Meghan, Amy Rose
...................................................
About  ln(abs(x/(x-1)))    << which needed one more set of ()

You should start by analyzing:
 x
-----
x - 1

You should work out that it is  > 0 [positive] everywhere except between 0 and 1.  (solution upon request)

So there is no real difficulty at x = -1.  (Naturally,  x = 1 and x = 0 are bad news.)

At  x = -1:   x/(x-1) is positive, so

abs(x/(x-1)) is just  x/(x-1), so you can go ahead and differentiate it as:

f(x) = ln(x/(x-1)) = ln x - ln(x-1)

f'(x) = 1/x - 1/(x-1),  and

f'(-1) = 1/(-1) - 1/(-1-1) =  oh, you can work out all those minuses -- I hate them, too.

............................


About inverses:  You asked:

Can you find an inverse function of an absolute value function?

Well, yes and no.  If you are interested in f-inv around a particular point, absolutely yes.
If you want to determine if  f-inv exists for all x as a function, then not always.

So, if you limit x to x NOT BETWEEN 0 AND 1, then you write:


y = ln(x/(x-1))

and if you limit x to x YES BETWEEN 0 AND 1, then you write:

y = ln(-x/(x-1))

and in either case do your standard 'find-the-inverse' routine.  (Again, solution on request.)

...............................



You wrote:


f(x)=((abs(x)-2)/(x-2)?   << too many () this time. [I know -- it's hard to see them on the computer.]


YOU ARE SURE YOU DID NOT MEAN:  f(x)=(abs(x-2))/(x-2), RIGHT?

We know it is -1 < x <= 1

OK, do the same thing -- Analyze it:

For  x > 0:

f(x)=(x-2)/(x-2) = 1.  So the range on the right is just one number: y = 1.


For  x < 0:
                      x + 2
f(x)=(-x-2)/(x-2) =  - ------
                      x - 2

Now you can just analyze that function.  You will find:

as x --> minus infinity,  f(x) --> -1

F(0) = - 2/ -2 = 1
         (x-2)(1) - (x+2)(1)
f'(x) = - --------------------
            (x - 2)^2

         x - 2 - x - 2
f'(x) = - ---------------
            (x - 2)^2

         + 4
f'(x) = -----------
        (x - 2)^2

which is ALWAYS POSITIVE.  

So the graph always rises :

from f(-inf) "=" -1 (the horizontal asymptote) to f(0) = +1.

I think that does it.

Let me know if you need more elucidation.
Let me know if you don't know what elucidation is.
Let me know if you need to spell elucidation.