How do you find #f^-1(x)# given #f(x)=(x+2)^2+6#?

1 Answer
George C.
Jun 13, 2017

#f(x)# has no inverse function, but...

Explanation:

Given:

#f(x) = (x+2)^2+6#

We can attempt to derive an inverse function as follows:

Let:

#y = f(x) = (x+2)^2+6#

Subtract #6# from both ends to get:

#y-6 = (x+2)^2#

Take the square root of both sides, allowing for both possible signs:

#+-sqrt(y-6) = x+2#

Transpose and subtract #2# from both sides to get:

#x = -2+-sqrt(y-6)#

Note that for any value of #y# (apart from #6#) in the range #[6, oo)# of #f(x)#, this formula gives us two possible values of #x#.

Since #f(x)# is many to one on its domain #(-oo, oo)#, its inverse is not a function, but a relation. So we cannot define #f^(-1)(y)# unless we restrict the range.

Alternatively, we can define an inverse image function #F^(-1)(y)# from #[6, oo)# to #2^RR# (the set of subsets of #RR#) by:

#F^(-1)(y) = { -2+sqrt(y-6), -2-sqrt(y-6) }#

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