Question

Please help asap find f -1

(picture of equation below)

 A.  f­-1 (x) = x2 +25; y>25

 
B.  f ­-1 (x) = x2 + 5; y > 5  
C. f ­-1 (x) = (x + 5)2; y  >5  
D. f­-1 (x) = x2 +25; y> 5

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)= √(x-5)`,

the domain of f, is the set of all x for which
`√(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)= √(x-5)`

to find the inverse function
`f^(-1)`,

1. write f(x) as y:


`y= √(x-5)`

2. write x in terms of y:


`y= √(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