determine the largest set D of real numbers for which f : D => R is a function:

any ideas on how to do this practice question? cheers.

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- Mar 8th 2009, 10:34 PMjvignaciodetermining the largest set D of real numbers
determine the largest set D of real numbers for which f : D => R is a function:

any ideas on how to do this practice question? cheers. - Mar 8th 2009, 11:30 PMADARSH
X should be in the set D (called Domain)

Thus all x which give you a real y are in this domain as co-domain is R

Solve this inequality

......so that we don't have negative inside root

And obviously

...since 0 in denominator is not allowed

Thus every x which satisfies the above two conditions is your answer - Mar 8th 2009, 11:53 PMjvignacio
- Mar 9th 2009, 01:02 AMjvignacio
is this true for

?? - Mar 9th 2009, 01:33 AMADARSH
- Mar 11th 2009, 09:34 PMADARSH
I think you are still in doubt , isn't it , I saw you peeking at this again and again(Giggle)

---Now if the denominator on LHS is positive

Hence but in the initial condition we have taken it greater than 6

Hence this is correct for

---Now when the denominator on LHS is negative

So there will be a change in inequality sign when we will multiply it by

(x-6) on both sides

Hence

but this is against initial condtion hence no answer in this range of x

----When the**denominator**on LHS is negative

So on multiplying the sign of inequality changes

Hence

thus is an answer

Now for the complete answer we have to take a combination of all these solution sets(answer as in my last post) (Nod)

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I am not sure if you were unclear about it, but I think if you were this will be helpful ,ask incase if you still find any trouble*(remember the change of inequality sign on multiplying with a negative number on both sides)*

(Thinking) - Mar 11th 2009, 09:38 PMjvignacio