Another Continuity Problem

**Afterme** · September 9th 2008, 04:15 PM

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

**Plato** · September 9th 2008, 04:21 PM

That function is meanless as posted.
Please review what you have posted and correct the function.
There is no x in the agrument.

**skeeter** · September 9th 2008, 04:29 PM

is this your function?

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

**Plato** · September 9th 2008, 04:41 PM

In either case $f(x) = \sqrt {2\cos (x)} - \sqrt 3 \,\mbox{or} \,f(x) = \sqrt {2\cos (x) - \sqrt 3 }$ the function is not even define.
Can you have the square root of a negative number?

**Afterme** · September 9th 2008, 05:28 PM

Well there is an answer for it and it's

[0, pi/6] U [11pi/6, 2pi]

**skeeter** · September 9th 2008, 05:54 PM

I'm assuming that the function is

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

if that is the case, then

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

you know why, correct?

simplifying the inequality ...

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

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

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