domain of f(x)

All Real x.

See man111's great post below!

pls explain

\hspace{-16}\mathbf{(1)\;\;:: f(x)=\sqrt{x^{12}-x^9+x^4-x+1}}$\\\\ Now function $\mathbf{f(x)}$ is defined only when $\mathbf{x^{12}-x^9+x^4-x+1>=0}$\\\\ Now Let $\mathbf{g(x)=x^{12}-x^9+x^4-x+1}$\\\\ \textbf{case{(a)::\;}}If $\mathbf{x\leq 0\;,}$ Then $\mathbf{g(x)>0}$\\\\ \textbf{case{(b)::\;}}If $\mathbf{0<x<1 \;,}$ Then $\mathbf{g(x)=x^{12}+x^4-x^9+1-x=x^{12}+x^4(1-x^5)+(1-x)>0}$\\\\ \textbf{case{(c)::\;}}If $\mathbf{x\geq 1 \;,}$ Then $\mathbf{g(x)=x^{12}-x^9+x^4-x+1=x^{9}(x^3-1)+x(x^3-1)+1>0}$\\\\ So $\mathbf{g(x)>0\forall x\in\mathbb{R}}$\\\\ So Domain is $\mathbf{x\in\mathbb{R}}$

i didn't understand case (a) :(

If x=0 , Then g(x) = 1 >0

and If x<0, Then x^12>0, -x^9>0, x^4>0, -x>0, 1>0

So g(x) > 0