Well, I'm not done yet. Am I on the right track?

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- Jan 30th 2009, 06:02 PMnathan02079Problem with inequalities
Well, I'm not done yet. Am I on the right track?

- Jan 30th 2009, 06:33 PMmollymcf2009
Hi Nathan02079!

Are you supposed to be solving the inequality for x? Or just evaluating the function until you find what is ? - Jan 30th 2009, 06:34 PMnathan02079
I am solving for x.

- Jan 30th 2009, 07:00 PMmollymcf2009
Looks good to me!

- Feb 1st 2009, 05:58 AMpankaj
**NOT**yet

First of all you need to find the values of x for which all the square roots are possible.i.e quantities under the square root signs must be non-negative.

On taking the intersection of the above conditions you will get

which can be said to be the domain of the inequality.

Now the solutions you obtain by solving

i.e

i.e

Therefore the values of x satisfying both the inequalities are

which must satisfy .

Therefore the answer to the given question must be [5,7] - Feb 1st 2009, 08:06 AMvg284
i think answer is [4 7].

- Feb 1st 2009, 08:32 AMpankaj
Take and put it in ,you will obtain which is impossible since is not possible in real numbers

- Feb 8th 2009, 11:26 AMekledes
Therefore the answer to the given question must be {4}u[5,7]

- Feb 8th 2009, 03:43 PMpankaj
That's correct.