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The function \(f(x)=\frac{x-1}x\) is defined on the domain \((-\infty,0)\cup(0,\infty)\). Is \(f(x)\) invertible? If an inverse function exists, find a formula for the inverse and state its domain and range. If an inverse does not exist, explain why not.

Guest Jun 21, 2017
 #1
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This function is invertible since it's one-to-one

 

To find the inverse......write as

 

y  = (x -1) / x       multiply  both sides by x

 

xy = x - 1           subtract x from both sides

 

xy - x  = -1        factor the left side

 

x ( y - 1)   = -1      divide both sides by (y - 1)

 

x =  -1  / (y - 1)     and we can write

 

x = 1 / ( 1 - y)      "swap" x and y

 

y =  1 / (1 - x)     and for y, write  f-1 (x)

 

f-1(x)  =  1 / (1 - x)

 

The domain for this function will be (-inf, 1) U ( 1, inf)

The  range  will be  (-inf, 0) U (0, inf )

 

 

cool cool cool

CPhill  Jun 21, 2017

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