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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?

Note by Geneveve Tudence 1 year, 9 months ago

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Whenever we are calling it function domain is R+ and range too

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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.

Note that if you are using the radical "symbol" then you are talking only about the positive root. So, \(\sqrt{64} = 8\) and not \(\pm8\).

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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TopNewestWhenever we are calling it function domain is R+ and range too

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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.

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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 \(\pm8\).

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