How do you find the inverse of #f(x)=(x+3)^(1/2) + 2# and is it a function?

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Answer 1 · 2016-05-30T07:46:13

#f^(-1)(y) = (y-2)^2-3#

#f(x) = (x+3)^(1/2)+2#

Assuming that we are dealing with Real square roots, the (implicit) domain of this function is #[-3, oo)#.

Let #y = (x+3)^(1/2)+2#

Rearrange this equation so that #x# is isolated on one side.

First subtract #2# from both sides to get:

#y-2 = (x+3)^(1/2)#

Square both sides to get:

#(y-2)^2 = x+3#

Subtract #3# from both sides and transpose to get:

#x = (y-2)^2-3#

For any Real value of #y#, this formula gives us a unique value of #x#. So this is a suitable inverse function:

#f^(-1)(y) = (y-2)^2-3#

The domain of this inverse function is the whole of #RR#, i.e. #(-oo, oo)# and its range is #[-3, oo)#, the domain of #f(x)#.

Footnote

Note that though #f^(-1)(y)# is an inverse function for #f(x)#, #f(x)# is not an inverse function for #f^(-1)(y)#.

The function #f^(-1)(y)# is not one-one and we find (for example):

#f(f^(-1)(color(blue)(1))) = f(-2) = color(blue)(3)#