<s>[INST] Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be? I tried solving this question by 
$1.$ $-1$ and $4$ will not be in domain because denominator can not be zero . 
$2.$ Either both denominator and numerator will be positive or negative so that whole term in root becomes positive.
But I am not able to solve the upper part , can it be done by taking $\log$ ? 
 [/INST] The zeroes of the denominator are $\frac{4 \pm \sqrt{32}}{2} = 2(1 \pm \sqrt{2})$. 
Therefore the denominator is negative iff $2(1 - \sqrt{2}) < x < 2(1 + \sqrt{2})$.
The upper part is positive if $x<0$, zero if $x=0$, and negative is $x>0$
So they ae both positive if $x < 2(1 - \sqrt{2})$.
They are both negative if $0 \leq x < 2(1 + \sqrt{2})$.
The answer is therefore $(\leftarrow, 2(1 - \sqrt{2})) \cup [0, 2(1 + \sqrt{2}))$
</s>