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 $\log$ ? 

To obtain the domain of $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$, the expression inside the radical must be positive and the denominator of the expression must be nonzero.
The numerator $3^ x - 4^x$ is positive when $x < 0$. The denominator $x^2-4x-4$ is positive when $x > 2(1+ \sqrt{2})$ or when $x < 2 (1 - \sqrt{2})$. Hence, the expression inside the radical sign is positive when $x < 2 (1 - \sqrt{2})$ or when $0 <x < 2(2 + \sqrt{2})$.
The values where the denominator are zero are $2(1 \pm \sqrt{2})$. They are not in the stated intervals above. 
Thus the domain (in interval notation) is $(- \infty,2 (1 - \sqrt{2})) \cup (0,2 (1 + \sqrt{2}))$.
