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.
What's the range of this function?
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algebra-precalculus
functions
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0The 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. – 2011-10-16
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0Yes, I did confuse. – 2011-10-16
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9Instead of editing the question title to add "Solved", you could answer your own question... :) – 2011-10-16
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1@okelvin: In fact, please do as J.M. suggests: write out your solution as an answer. Then you can later even accept it! – 2011-10-16
1 Answers
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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.