Expert: Paul Klarreich - 8/20/2009

Question
Hi, I have an assignment for class and it requires me to answer the following question:

Solve the following:

(a) Find the inverse of y = f(x) = (1 + e^(-x))/(1 - e^(-x))
(b) Find the domain of y = f(x) = [(1/(x-1)) + (3/2x)]^(1/2) + ln (2-x)

Can you please help me?

Answer
Questioner: Sana
Country: Canada
Category: Calculus
Private: No
Subject: Calculus
Question: Hi, I have an assignment for class and it requires me to answer the following question:

Solve the following:

(a) Find the inverse of y = f(x) = (1 + e^(-x))/(1 - e^(-x))
(b) Find the domain of y = f(x) = [(1/(x-1)) + (3/2x)]^(1/2) + ln (2-x)

Can you please help me?
.......................................
To find an inverse of f(x).

1. Write y = f(x)

y = (1 + e^(-x))/(1 - e^(-x))

2. Swap the letters x,y:

x = (1 + e^(-y))/(1 - e^(-y))

3. Solve for y.

   1 + e^(-y)   e^y
x = -----------  ----   << clear
   1 - e^(-y)   e^y

   e^y + 1
x = -------
   e^y - 1

Write  u = e^y (just for convenience)

   u + 1
x = -----
   u - 1

Solve for u:

ux - x = u + 1
ux - u = x + 1
u(x - 1) = x + 1
   x + 1
u = -----
   x - 1

UNWRITE  u = e^y

     x + 1
e^y = -----
     x - 1

y = ln(....) and this is your  f-inv(x)

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



f(x) = [(1/(x-1)) + (3/2x)]^(1/2) + ln (2-x)

Exclude any 'bad' values for x, such as:

-- make a denominator zero.
So you cannot have  x = 1 or x = 0
-- make a sqrt(negative)
So you cannot have  1/(x-1) + 3/2x < 0

Solve it:
 1      3
----- + --- < 0
x - 1   2x

 1       -3
----- <  ---
x - 1     2x

Now it gets a bit complicated, so I'll leave this part to you.
Suggestion: Assume  x is negative.

-- make a ln(zero or negative)

So you cannot have  2 - x <= 0
or  2 <= x, which reads better as  x >= 2