The document discusses finding the inverse function of f, outlining its domain and range. The original function f(x) = 1/2 x^2 - 1 has a domain of (-inf, +inf) and a range of [-1, +inf). The inverse function f^-1 is determined to be sqrt(7x + 7, with a domain of [-1, +inf) and a range of [0, +inf).

1. Consider the following function. Find the inverse function State the domain and range of f.
(Enter your answers using interval notation.)
Solution
Fisrt part is correct ie. F^-1 = sqrt(7x +7)
domain and range of function f (x) = 1/2 x^2 - 1 is all values of x greater than equal to 0
domain ( -inf, +inf)
Range : find Y intercpt by plugging x=0 we get y =-1
[ -1, + inf)
Domain of inverse function: Check where function does not exist: 7x+7 >=0
x> = -1
Domain:: [ -1 , inf)
Range : f(x) exist for >=0
So, [0, inf)