determining 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.

Quote:

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Originally Posted by jvignacio

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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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

Quote:

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Originally Posted by ADARSH

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

---

thanks man! appreciated

is this true for

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

??

Quote:

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Originally Posted by jvignacio

is this true for

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

??

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DO you think its correct for 5.5

Do you think its incorrect for 4


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

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)

Quote:

---

Originally Posted by ADARSH

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)

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

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)

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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