# determining the largest set D of real numbers

• Mar 8th 2009, 11:34 PM
jvignacio
determining the largest set D of real numbers
determine the largest set D of real numbers for which f : D => R is a function:

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

any ideas on how to do this practice question? cheers.
• Mar 9th 2009, 12:30 AM
Quote:

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

$h(x) = \sqrt{\frac{1}{x-6}+1}$
any ideas on how to do this practice question? cheers.

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

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

And obviously $x \ne 6$

...since 0 in denominator is not allowed
Thus every x which satisfies the above two conditions is your answer
• Mar 9th 2009, 12:53 AM
jvignacio
Quote:

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

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

And obviously $x \ne 6$

...since 0 in denominator is not allowed
Thus every x which satisfies the above two conditions is your answer

thanks man! appreciated
• Mar 9th 2009, 02:02 AM
jvignacio
is this true for

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

??
• Mar 9th 2009, 02:33 AM
Quote:

Originally Posted by jvignacio
is this true for

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

??

DO you think its correct for 5.5

Do you think its incorrect for 4

(Smile) Answer is (-infinity , 5) And (6, infinity)
• Mar 11th 2009, 10:34 PM
I think you are still in doubt , isn't it , I saw you peeking at this again and again(Giggle)

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

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

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

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

Hence this is correct for $x>6$

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

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

----When $x \le 5$ the denominator on LHS is negative
So on multiplying the sign of inequality changes

Hence $x-5 \le 0$

thus $x \le 5$ 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)

------------------------------------

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, 10:38 PM
jvignacio
Quote:

I think you are still in doubt , isn't it , I saw you peeking at this again and again(Giggle)

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

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

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

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

Hence this is correct for $x>6$

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

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

----When $x \le 5$ the denominator on LHS is negative
So on multiplying the sign of inequality changes

Hence $x-5 \le 0$

thus $x \le 5$ 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)

------------------------------------

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)

haha thanks for that, yeah ive been trying to study it and really try understand it. :) cheers for the last post. made things more simple