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Re: If #x is defined for all x > –2 as the square root of the nu
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11 Aug 2018, 16:35
Explanation
This function defines a made-up symbol, rather than using traditional notation such as f(x).
First, translate the function:
\(#x = \sqrt{x+2}\)
The square root of any value greater than or equal to 0 is the non-negative square root of the value.
That is, the square root of 4 is just +2, not –2. Thus:
\(#7 =\sqrt{7+2} =\sqrt{9} = 3\)
\(#(-1) = \sqrt{-1+2}= \sqrt{1}= 1\)
Finally, \(#7 - #(-1) = 3 - 1 = 2\).