# Thread: Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5

1. ## Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5

Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
What is the domain of f-1?

-9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5

2. Originally Posted by azninvasion
Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
What is the domain of f-1?

-9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5
the domain and range of a function and it's inverse are "swapped".

3. so, does that mean the answer is -4 ≤ x ≤ 5????

4. Originally Posted by azninvasion
so, does that mean the answer is -4 ≤ x ≤ 5????
No.

The domain of $\displaystyle f^{-1}$ equals the range of $\displaystyle f$

5. Originally Posted by e^(i*pi)
No.

The domain of $\displaystyle f^{-1}$ equals the range of $\displaystyle f$
-4 ≤ f-1(x) ≤ 5 Its this then?

7. 0 ≤ x ≤ 9 maybe its this? haha this is like the hardest math problem on this forum ><

8. Originally Posted by azninvasion
Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
What is the domain of f-1?

-9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5
domain are x-values ...

since the range of $\displaystyle f(x)$ is $\displaystyle 0 \le f(x) \le 9$ , the domain of $\displaystyle f^{-1}(x)$ is $\displaystyle 0 \le x \le 9$

9. Originally Posted by skeeter
domain are x-values ...

since the range of $\displaystyle f(x)$ is $\displaystyle 0 \le f(x) \le 9$ , the domain of $\displaystyle f^{-1}(x)$ is $\displaystyle 0 \le x \le 9$
oooooo, why didn't you say so in the first place. now I get it.

+thanked.

10. Originally Posted by skeeter
the domain and range of a function and it's inverse are "swapped".
... that's what I said "in the first place".

11. Originally Posted by skeeter
... that's what I said "in the first place".
but your post was in the second place

not that funny now that i read it