Question

Find the domain of definition of the following function:
$f\left(x\right)=\sqrt{16x-x^5}+{\log }_{\frac{1}{2}}(x^2-1)$

Answer 1

$16x-x^5\ge 0x^5-16x\le 0x(x-2)(x+2)(x^2+4)\le 0x\in (-\infty ,-2]\cup [0,2]$

And

${\log }_{0.5}(x^2-1)x^2-1>0(x+1)(x-1)>0x\in (-\infty ,-1)\cup (1,\infty )$

Taking intersection we get

$x\in (-\infty ,-2]\cup (1,2]$