Another continuity problem I'm having trouble on

On the interval [0,2pi] where is the function f(x) = (sqrt 2cosx - (sqrt 3)) continuous?

Thanks again guys

Edit : Sorry about that Plato. Forgot to put the X after Cos. Thanks

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- Sep 9th 2008, 03:15 PMAftermeAnother Continuity Problem
Another continuity problem I'm having trouble on

On the interval [0,2pi] where is the function f(x) = (sqrt 2cosx - (sqrt 3)) continuous?

Thanks again guys

Edit : Sorry about that Plato. Forgot to put the X after Cos. Thanks - Sep 9th 2008, 03:21 PMPlato
That function is meanless as posted.

Please review what you have posted and correct the function.

There is no x in the agrument. - Sep 9th 2008, 03:29 PMskeeter
is this your function?

$\displaystyle f(x) = \sqrt{2\cos{x} - \sqrt{3}}$ - Sep 9th 2008, 03:41 PMPlato
In either case $\displaystyle f(x) = \sqrt {2\cos (x)} - \sqrt 3 \,\mbox{or} \,f(x) = \sqrt {2\cos (x) - \sqrt 3 } $ the function is not even define.

Can you have the square root of a negative number? - Sep 9th 2008, 04:28 PMAfterme
Well there is an answer for it and it's

[0, pi/6] U [11pi/6, 2pi] - Sep 9th 2008, 04:54 PMskeeter
I'm assuming that the function is

$\displaystyle f(x) = \sqrt{2\cos{x} - \sqrt{3}}$

if that is the case, then

$\displaystyle 2\cos{x} - \sqrt{3} \geq 0$

you know why, correct?

simplifying the inequality ...

$\displaystyle \cos{x} \geq \frac{\sqrt{3}}{2}$

your "answer" is the set of x-values for which the above inequality is true.