# determining the largest set D of real numbers

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• March 12th 2009, 02:12 AM
jvignacio
determining the largest set D of real numbers
hi guys i got another example here which im not to confident with..

f : D -> R
$
h(x) = \frac{1}{\sqrt{x}-6}+1$

im thinking this would have a slightly different approach because not the whole thing is getting squared.
• March 12th 2009, 02:19 AM
ADARSH
Quote:

Originally Posted by jvignacio
hi guys i got another example here which im not to confident with..

f : D -> R
$
h(x) = \frac{1}{\sqrt{x}-6}+1$

im thinking this would have a slightly different approach because not the whole thing is getting squared.

Since denominator can't be zero x is not equal to 36

Since negative inside square root is not allowed

x (>=) 0

these are the only two conditions
• March 12th 2009, 02:31 AM
jvignacio
Quote:

Originally Posted by ADARSH
Since denominator can't be zero x is not equal to 36

Since negative inside square root is not allowed

x (>=) 0

these are the only two conditions

yes yes thats right. so x has to be greater than 0 because u cant square a negative number and if x was 36, then 6-6 = 0 and u cant divide by a 0. UNDERSTOOD :)

x >=0 and x not equal to 36