Inverse

Problem:



Solution:

Part (a)
$ f^{-1}(x) = \sqrt{15 + x} $

Part (b)

Part (c)

The domain of $ f $ = Range of $ f^{-1} $ = $ [0, +\infty) $.
The range of $ f $ = domain of $ f^{-1} $ = $ [-15, +\infty) $.