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# Inverse functions

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(a) Let $$f(x) : (-\infty,0) \cup (0,\infty) \to \mathbb{R}$$ be defined by $$f(x) = x - \frac{1}{x}.$$
Show that f(x) has no inverse function.
(b) Let $$g(x) : (0,\infty) \to \mathbb{R}$$ be defined by $$g(x) = x - \frac{1}{x}.$$
Show that g(x) has an inverse function.

I'm a little confused on how to find the inverse function of f(x). I thought I could prove that two x values could both have the same y value, but I'm not really sure how. Solving for the inverse of f(x) also confused me a little. How should I proceed?

Thanks!

Dec 30, 2020

### 2+0 Answers

#1
+1

y  = x   - 1/x

y =  [ x^2 -1 ] / x

xy   =  x^2    - 1

Trying to get x by itself  and then "swap" x  and  y is sticky

Here's one approach   by completing the  square on  x

yx  =  x^2  - 1          rearrange as

x^2  - yx     - 1    =   0       complete the square

Take 1/2  of y  =  y/2   square it  = y^2/4   add to both  sides

x^2  - yx   + (y^2/4)   =  1  + (y^2/4)       the left side  factors as

( x - y/2)^2  = ( 4 + y^2) / 4

x - y/2  =  ± √(4 + y^2)  / 2

x =  ± √(4 + y^2)  /  2  + y/2

x  =   [ y ± √ (4 + y^2) ]   / 2      "swap"  x and  y

y =  [ x ± √ (4 + x^2 ) ]  / 2   =    the "inverse"

Note  that  if  x  =  1   in the  first function, then  y   =  0

So   the point    ( 0, 1)   should  be  on the  "inverse" graph...so

1  =  [ 0 ± √ [ 4 + 0  ]   ] / 2

1  = ± √4/2

1  =  ±1

This is  only  true  if we  take the positive root

So.....  x  - 1/x     has no inverse   on   (-inf, 0)  U (0, inf)  because  we don't get a unique point for the "inverse"

However....if we restrict the domain of  g (x) to    (0, inf)    then  the inverse  becomes

y =  [ x +  √ [ 4  + x^2 ]   ]  /  2

Now.....note if   x =   2  in  g(x)   then   g(2)  =  3/2

And the point  ( 2,3/2) is  on g

So....the point   (3/2 , 2)   should be  on the  inverse   of  g

So

2 =   [ 3/2 +  √ [ 4 + (3/2)^2 ]   ]/  2

4 =  3/2  +    √ [ 4 + 1.5^2 ]

4 =  3/2  +  √ [ 4 + 2.25  ]

5/2  =  √(6.25

2.5   = 2.5      which is true

So we  have an inverse if we  restrict the  domain  to  (0, inf)

Also.....look at the  graph here  for  x  -1/x

https://www.desmos.com/calculator/g4o70vl223

This  graph  does not pass the horizontal line test  ( a requirement for an inverse)

However....if we  restrict the  domain  to   (0, inf)   we will have a  graph that will pass the horizontal line test,  and so, it will have an inverse   Dec 30, 2020
edited by CPhill  Dec 30, 2020
#2
+1

That helped a lot, getting the inverse was really tricky for me but I understand now. Thank you for the help!

Guest Dec 31, 2020