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Let \(f(x) = x^2 + 4x + 5\) for \(x \le -2.\) Find \(f^{-1}(x).\)

 Feb 20, 2020
 #1
avatar+111433 
+1

This one is a little tricky, DragonLord  !!!

 

The goal is to get x by itself and then "swap" x  and y

 

We can write

 

y  = x^2  + 4x  +  5      subtract  5  from both sides

 

y - 5 =  x^2  + 4x         complete  the  square  on x

 

y  -5  + 4  =  x^2  + 4x  +  4    factor the right side.....simplify  the left

 

y - 1  = (x + 2)^2      take both roots

 

 ±√[y - 1 ]  = x + 2     subtract 2 from  both sides

 

x = ±√[y -1 ] - 2   "swap"  x and  y

 

y  = ±√ [x -1] - 2       and we need to take the negative value  [I'll explain this in a second]

 

y = - √ [x -1]   -2    this is the inverse

 

However.....we have a resticted domain

 

To find  out what the  new  domain is we need to  solve  this : 

 

(-2)^2  + 4(-2)  +  5  =

 

4  - 8  + 5

 

-4 + 5

 

1

 

 

This means that the  new  domain is   x ≥  1

 

 

See the graph here  :  https://www.desmos.com/calculator/xeind7czif

 

Here's why  we need  the   "-"  value on the root  function of the  inverse

 

Note  that  the  point   (-4, 5)  i on the origimal graph

 

The inverse  "switches"  these coordinates...so....the  point  ( 5, -4)  must  be  on the inverse

 

See the graph again to confirm this  : https://www.desmos.com/calculator/kt9chuswm0

 

cool cool cool

 Feb 20, 2020
edited by CPhill  Feb 20, 2020
edited by CPhill  Feb 21, 2020
 #2
avatar+278 
+1

Very precise strategy to find the inverse. Thank you for your help <3

 Feb 21, 2020

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