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

• November 15th 2009, 01:15 PM
azninvasion
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 (Headbang)
• November 15th 2009, 01:25 PM
skeeter
Quote:

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 (Headbang)

the domain and range of a function and it's inverse are "swapped".
• November 15th 2009, 01:44 PM
azninvasion
so, does that mean the answer is -4 ≤ x ≤ 5????
• November 15th 2009, 01:59 PM
e^(i*pi)
Quote:

Originally Posted by azninvasion
so, does that mean the answer is -4 ≤ x ≤ 5????

No.

The domain of $f^{-1}$ equals the range of $f$
• November 15th 2009, 02:26 PM
azninvasion
Quote:

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

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

-4 ≤ f-1(x) ≤ 5 Its this then?
• November 15th 2009, 04:00 PM
eliteplague
• November 15th 2009, 04:03 PM
azninvasion
0 ≤ x ≤ 9 maybe its this? haha this is like the hardest math problem on this forum ><
• November 15th 2009, 04:19 PM
skeeter
Quote:

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 $f(x)$ is $0 \le f(x) \le 9$ , the domain of $f^{-1}(x)$ is $0 \le x \le 9$
• November 15th 2009, 04:24 PM
azninvasion
Quote:

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

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

oooooo, why didn't you say so in the first place. now I get it.

+thanked.
• November 15th 2009, 04:34 PM
skeeter
Quote:

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".
• November 15th 2009, 06:04 PM
Krahl
Quote:

Originally Posted by skeeter
... that's what I said "in the first place".

but your post was in the second place (Rofl)

not that funny now that i read it :(