The inverse of a function #f# is a function #f^(-1)# such that #f^(-1)(f(x)) = x# for every #x# in the domain of #f#.
One trick for finding the inverse of a function #f# is to set #x=f(y)# and then solve for #y#. The resulting equation will be of the form #y=f^(-1)(x)#.
If we apply that trick to the given function, though, we run into an issue:
Let #x = h(y) = 4-y^2#
#=> y^2 = 4-x#
#=> y = +-sqrt(4-x)#
It seems, then, that we would have #h^(-1)(x) = +-sqrt(4-x)#. Unfortunately, this is not a function, as by definition, a function can only produce one value for any given input. Here we rely on the additional information given in the problem.
We are given that #x<=0#, meaning that #h^(-1)(h(x)) <= 0#. This tells us that the inverse function should be producing negative values, meaning we want #-sqrt(4-x)#, rather than #sqrt(4-x)#. Thus, our final answer is
#h^(-1)(x) = -sqrt(4-x)#
(As a side note, we do not have to worry about taking the square root of a negative number as the domain of #h^(-1)# is the same as the range of #h#, which is #(-oo, 4]#. Thus we will not apply #h^(-1)# to any value greater than #4#.)
