Thread: determining the largest set D of real numbers

1. 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.

2. 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

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

4. is this true for

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

??

5. 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

Answer is (-infinity , 5) And (6, infinity)

6. I think you are still in doubt , isn't it , I saw you peeking at this again and again

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

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

---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)

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

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)

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

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

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

---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)

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

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)

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