+304 Let \(f(x) = x^2 + 4x + 5\) for \(x \le -2.\) Find \(f^{-1}(x).\)
+128407 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
+304 Very precise strategy to find the inverse. Thank you for your help <3
