I.
Given a function f with domain D and range R, and the inverse function
[tex]f ^{-1} [/tex],
then the domain of [tex]f ^{-1} [/tex] is the range of f, and the range of [tex]f ^{-1} [/tex] is the domain of f.
II. We are given the function [tex]f(x)= \sqrt{x-5} [/tex],
the domain of f, is the set of all x for which [tex]\sqrt{x-5} [/tex] makes sense, so x is any x for which x-5[tex] \geq [/tex]0, that is x≥5.
the range is the set of all values that f can take. Since f is a radical function, it never produces negative values, in fact in can produce any value ≥0
Thus the Domain of f is [5, ∞) and the Range is [0, ∞)
then , the Domain of [tex]f ^{-1} [/tex] is [0, ∞) and the Range of [tex]f ^{-1} [/tex] is [5, ∞)
III.
Consider [tex]f(x)= \sqrt{x-5} [/tex]
to find the inverse function [tex]f^{-1} [/tex],
1. write f(x) as y:
[tex]y= \sqrt{x-5} [/tex]
2. write x in terms of y:
[tex]y= \sqrt{x-5}[/tex]
take the square of both sides
[tex]y^{2} =x-5[/tex]
add 5 to both sides
[tex]y^{2} +5=x[/tex]
[tex]x=y^{2} +5[/tex]
3. substitute y with x, and x with [tex]f^{-1}(x) [/tex]:
[tex]f^{-1}(x)=x^{2} +5[/tex]
These steps can be applied any time we want to find the inverse function.
IV. Answer:
[tex]f^{-1}(x)=x^{2} +5[/tex], x≥0
y≥0, where y are all the values that [tex]f^{-1}[/tex] can take
Remark: the closest choice is B
