Question 1093996: Let f(x) = (x-8)^2
Find a domain on which f is one to one and non decreasing
Answer= [8,oo) BUT
What is the inverse of f restricted to this domain??
f^-1(x)=?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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:
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.
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