Question 1093996
set y = (8-x)^2


replace y with x and x with y to get:


x = (8-y)^2


solve for y as follows:


take square root of both sides to get plus or minus sqrt(x)= 8-y.


this give you two equations.


plus sqrt(x) = 8 - y.


minus sqrt(x) = 8 - y


add y to both sides of the first equation and subtract sqrt(x) from both sides of the first equation to get:


y = 8 - sqrt(x)


add y to both sides of the second equation and add sqrt(x) to both sides of the second equation to get:


y = 8 + sqrt(x)


these are your inverse equations.


you need to apply the test for inverse euation to see which is inverse for what.


when x > 8, then if f(x) = (x,y), then f^-1)(x) = (y,x)


so take a value of x greater than 8, say 12, and solve for y = (8-x)^2.


you will get y = (8-12)^2 = (-4)^2 = 16.


your coordinate point of the regular function is (12,16).


now take y = 8 + sqrt(x) and replace x with 16 and solve for y.


you will get y = 8 + sqrt(16) = 8 + 4 = 12.


your coordinate point of the inverse function is (16,12).


y = 8 + sqrt(x) is your inverse function to the part of y = (8-x)^2 that is increasing.


graphically, it looks like this:


<img src = "http://theo.x10hosting.com/2017/091801.jpg" alt="$$$" >


the inverse function is a reflection about the line y = x.


(x,y) in the original graph is reflected by (y,x) in the inverse graph.


this is clearly seen in the graph.