Why do we call \(f(x) = \sqrt x\) a function, if by definition \(y = \sqrt x\) is not a function since for included values of \(x\), it gives us more than one value of \(y\) considering the positive and negative answers in the radicals?

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TopNewestWhenever we are calling it function domain is R+ and range too – Satyam Tripathi · 8 months, 3 weeks ago

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Have you read the list of common misconceptions wiki ? I recommend reading #8 in the "Powers and Square Roots" section of that wiki. – Prasun Biswas · 8 months, 3 weeks ago

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Note that if you are using the radical "symbol" then you are talking only about the positive root.

So, \(\sqrt{64} = 8\) and not \(\pm8\). – Yatin Khanna · 8 months, 3 weeks ago

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