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

$\displaystyle h(x) = \sqrt{\frac{1}{x-6}+1}$

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:

$\displaystyle h(x) = \sqrt{\frac{1}{x-6}+1}$

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

$\displaystyle \frac{1}{x-6}+1 \ge 0$......so that we don't have negative inside root

And obviously $\displaystyle x \ne 6$

...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

$\displaystyle \frac{1}{x-6}+1 \ge 0 = $$\displaystyle x \ge 5$

?? - 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)

$\displaystyle \frac{1}{x-6} +1 \ge 0$

$\displaystyle \frac{x-5}{x-6} \ge 0$

---Now if $\displaystyle x> 6 $the denominator on LHS is positive

Hence $\displaystyle x\ge 5$ but in the initial condition we have taken it greater than 6

Hence this is correct for $\displaystyle x>6$

---Now when $\displaystyle 5< x<6$ 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 $\displaystyle x-5 < 0 $

$\displaystyle

\implies x < 5$ but this is against initial condtion hence no answer in this range of x

----When $\displaystyle x \le 5$ the**denominator**on LHS is negative

So on multiplying the sign of inequality changes

Hence $\displaystyle x-5 \le 0 $

thus $\displaystyle x \le 5$ 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