Question

Help out with the domain of this function (see attachment)

My conditions:

The huge square root >= 0
The smaller square root at the numerator >= 0
1-cosx != 0
the whole denominator != 0
The small square root at the denominator >= 0

I find the solutions to be 2 14 years ago

Answer 1

sowwy, 0<x<2

Answer 2

yikes what a mess. one thing is for sure, x cannot be 0
and
\[2x^2-16x+28\] has to be positive , so x cannot be in the interval
\[(4-\sqrt{2},4+\sqrt{2})\]

Answer 3

and also you have
\[\sqrt{2x^2-16x+28}-x\geq0\] which means
\[x\leq 2, x\geq 14\]

Answer 4

last thing you have to worry about is that
\[\pi-\tan^{-1}(\sqrt{3^{\frac{1}{x}}-2^{\frac{1}{x}}})\neq 0\]

Answer 5

so
\[\tan^{-1}(\sqrt{3^{\frac{1}{x}}-2^{\frac{1}{x}}})\neq \frac{\pi}{2}\]
but
\[\frac{\pi}{2}\] is not in the range of arctangent, so i don't think this is a problem