# Range trick question?

• Feb 10th 2010, 03:16 PM
nox16
Range trick question?
I don't understand how the answer is "x is greater than or equal to 0." I already solved it algebraically but don't see how the answer is x > 0. Please show me how to get that answer.

Note: The book is not wrong, the teacher said it was a trick question and the answer is x is greater than or equal to 0. But we must show work on how to get that.

http://i46.tinypic.com/2h6y0sp.jpg

Thank you.
• Feb 10th 2010, 03:22 PM
icemanfan
In that function f(x), x can be any real number. So the domain of f(x) is all real numbers. The range of f(x) is also all real numbers. Which leaves me wondering what the question is.
• Feb 10th 2010, 03:26 PM
skeeter
Quote:

Originally Posted by nox16
I don't understand how the answer is "x is greater than or equal to 0." I already solved it algebraically but don't see how the answer is x > 0. Please show me how to get that answer.

Note: The book is not wrong, the teacher said it was a trick question and the answer is x is greater than or equal to 0. But we must show work on how to get that.

http://i46.tinypic.com/2h6y0sp.jpg

Thank you.

what was the question, again?
• Feb 10th 2010, 03:35 PM
satis
I think the question is, what's the range of $(x-4)^4$

The domain of a function is the range of its inverse, so we can take the inverse of $(x-4)^4$.

$x=(y-4)^3$
$\sqrt[3]{x} = y-4$
$y=\sqrt[3]{x} + 4$

The domain of that inverse is all real numbers... so that should be the range of the original. I'm not sure I see why the range should be > 0.
• Feb 10th 2010, 04:04 PM
nox16
Sorry for not being so clear about the problem.

I was referring to what is the range of that problem. In the back of the book it says x is greater than or equal to 0. However I got the same answer as satis did and don't fully understand why the answer is x > 0.

Same goes for this problem I got on the test. I got it wrong because I put range is all real numbers (y^2 + 3 = x) when the answer is "All real numbers greater than or equal to 0".
http://i50.tinypic.com/2lwkgwo.jpg
• Feb 10th 2010, 04:10 PM
skeeter
Quote:

Originally Posted by nox16
Sorry for not being so clear about the problem.

I was referring to what is the range of that problem. In the back of the book it says x is greater than or equal to 0. However I got the same answer as satis did and don't fully understand why the answer is x > 0.

Same goes for this problem I got on the test. I got it wrong because I put range is all real numbers (y^2 + 3 = x) when the answer is "All real numbers greater than or equal to 0".
http://i50.tinypic.com/2lwkgwo.jpg

what does the graph of $f(x) = \sqrt{x-3}$ tell you about the range ?
• Feb 10th 2010, 04:50 PM
nox16
Range is (3,∞). Oh I see how how the answer of range is x is greater than or equal to 0. If possible, how would you know that is the limit though. Would I need to graph it like you did, or could I have solved it mathematically solving for x in that equation f(x) = sqrt x-3.