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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