Question

Please help asap find f -1

Answer 1

I.
Given a function f with domain D and range R, and the inverse function 
$f ^{-1}$,

then the domain of $f ^{-1}$ is the range of f, and the range of $f ^{-1}$ is the domain of f.

II. We are given the function $f(x)= \sqrt{x-5}$,

the domain of f, is the set of all x for which $\sqrt{x-5}$ makes sense, so x is any x for which x-5$\geq$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 $f ^{-1}$ is [0, ∞) and the Range of $f ^{-1}$ is [5, ∞) 

III.
Consider $f(x)= \sqrt{x-5}$

to find the inverse function $f^{-1}$, 

1. write f(x) as y:

$y= \sqrt{x-5}$

2. write x in terms of y:

$y= \sqrt{x-5}$

take the square of both sides

$y^{2} =x-5$

add 5 to both sides

$y^{2} +5=x$

$x=y^{2} +5$

3. substitute y with x, and x with $f^{-1}(x)$:

$f^{-1}(x)=x^{2} +5$


These steps can be applied any time we want to find the inverse function.

IV. Answer: 

$f^{-1}(x)=x^{2} +5$, x≥0

y≥0, where y are all the values that $f^{-1}$ can take

Remark: the closest choice is B