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

**azninvasion** · November 15th 2009, 01:15 PM

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?

Would the answer be -9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5

**skeeter** · November 15th 2009, 01:25 PM

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?

Would the answer be -9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5

the domain and range of a function and it's inverse are "swapped".

**azninvasion** · November 15th 2009, 01:44 PM

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

**e^(i*pi)** · November 15th 2009, 01:59 PM

Originally Posted by azninvasion

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

No.

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

**azninvasion** · November 15th 2009, 02:26 PM

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?

**eliteplague** · November 15th 2009, 04:00 PM

I think its your first answer,

**azninvasion** · November 15th 2009, 04:03 PM

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

**skeeter** · November 15th 2009, 04:19 PM

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?

Would the answer be -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 $0 \le x \le 9$

**azninvasion** · November 15th 2009, 04:24 PM

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.

**skeeter** · November 15th 2009, 04:34 PM

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".

**Krahl** · November 15th 2009, 06:04 PM

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

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