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Domain is all real numbers. $f(x) = \sqrt{x - 3}$. The book's answer is "all non negative real numbers". But this includes all values between 0 and 3. Squareroot of negative number doesn't exist in real number set.

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The domain isn't all real numbers, as $\sqrt{-1 - 3} = 2i$ which is not real. Can you find say an $x$ such that $2 = \sqrt{x - 3}$? How about $-1 = \sqrt{x - 3}$? I think you're mixing up the definition of domain and range. –  user1205 Oct 16 '11 at 0:31
    
Yes, I did confuse. –  0 kelvin Oct 16 '11 at 0:40
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Instead of editing the question title to add "Solved", you could answer your own question... :) –  J. M. Oct 16 '11 at 0:41
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@okelvin: In fact, please do as J.M. suggests: write out your solution as an answer. Then you can later even accept it! –  Arturo Magidin Oct 16 '11 at 2:30
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1 Answer 1

Confused domain and image. Domain is [3, $+\infty$). Range is: what's the lowest value that the function can return? f(3) = 0, then the book's answer is right.

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