# Inverse FunctionWatch

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#1
Just a quick question:

Find the inverse of the function

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

thanks
0
14 years ago
#2
x/(1+x) for x>0
x/(1-x) for x<0
0
14 years ago
#3
Draw a graph!

Let x be such that |x| != 1. ["!=" means "not equal".]

For all y with |y| != 1,

x/(1 - |x|) = y
<=> x = y(1 - x) and x >= 0
____OR x = y(1 + x) and x <= 0
<=> x = y/(1 + y) and x >= 0
____OR x = y/(1 - y) and x <= 0

For all y with |y| > 1,

x/(1 - |x|) = y
<=> x = y/(1 + y) and x >= 0
____OR x = y/(1 - y) and x <= 0
<=> x = y/(1 + y)
____OR x = y/(1 - y)

For all y with 0 <= y < 1,

x/(1 - |x|) = y
<=> x = y/(1 + y) and x >= 0
____OR x = y/(1 - y) and x <= 0
<=> x = y/(1 + y)

For all y with -1 < y < 0,

x/(1 - |x|) = y
<=> x = y/(1 + y) and x >= 0
____OR x = y/(1 - y) and x <= 0
<=> x = y/(1 - y)

Finally we deal with y in {-1, 1}.

x/(1 - |x|) = 1
<=> x = (1 - x) and x >= 0
____OR x = (1 + x) and x <= 0
<=> x = 1/2

x/(1 - |x|) = -1
<=> x = -(1 - x) and x >= 0
____OR x = -(1 + x) and x <= 0
<=> x = -1/2

--

For all real y,

f^(-1)(y) =
__{ y/(1 + y) or y/(1 - y) ____ if |y| > 1
__{ y/(1 + y) ____ if 0 <= y < 1
__{ y/(1 - y) ____ if -1 < y < 0
__{ 1/2 ____ if y = 1
__{ -1/2 ____ if y = -1

Ie, for all real y,

f^(-1)(y) =
__{ y/(1 + y) or y/(1 - y) ____ if |y| > 1
__{ y/(1 + |y|) ____ if |y| <= 1

You might be able to simplify the proof by using the fact that f^(-1) has to be an odd function.

--

Check

f^(-1)(2) = 2/3 or -2
f^(-1)(1/2) = 1/3
f^(-1)(-1/5) = -1/6

But

f(2/3) = (2/3) / (1/3) = 2
f(-2) = -2/-1 = 2
f(1/3) = (1/3) / (2/3) = 1/2
f(-1/6) = -(1/6) / (5/6) = -1/5
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