Thread: Another 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

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

is this your function?

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

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?

Well there is an answer for it and it's

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

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.