Thread: determining 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.

Originally Posted by jvignacio

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.

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

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

......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

thanks man! appreciated

is this true for



??

Originally Posted by jvignacio

is this true for



??

DO you think its correct for 5.5

Do you think its incorrect for 4


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





---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)

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

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)

Originally Posted by ADARSH

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





---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)

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

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