<PRE><B>Question 756020</B>
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Your denominator does indeed exclude 0 and 2, but the argument of a square root function must be greater than or equal to zero.  Hence, the x cannot be less than 2 or the numerator is not defined.  So relying only on the numerator, you have x &gt;= 2, but since the denominator excludes x = 2, you are left with x &gt; 2.


And just an &quot;oh by the way&quot;, it is square rooT, not square rooF.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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