SOLUTION : $f(x)=\left(x^{12}-x^9+x^4-x+1\right)^{-1 / 2}$
Dr: $x^{12}-x^9+x^4-x+1>0$
For $x \leq 0$ it is obvious that for $f(x)$ to be defined $D r>0$.
For $x \geq 1,\left(x^{12}-x^9\right)+\left(x^4-x\right)+1$ is positive
Since $x^{12} \geq x^9, x^4 \geq x$.
For $0<x<1, \operatorname{Dr}=x^{12}+\left(x^4-x^9\right)+(1-x)>0$
Since $x^4>x^9, x<1$.
Hence $\mathrm{Dr}>0$ for all $x \in R$
Domain is $x \in R$
