Finding domain of $\sqrt{x^{12} - x^9 + x^4 - x + 1}$ How to find the domain of this function $\sqrt{x^{12} - x^9 + x^4 - x + 1}$?
Its really hard to find all 12 roots and plot them or use tricks like "wavy curve" to check where the expression in the square root is positive . But I thought (i'm not sure) if we can prove that the function is positive for all value then the job would be done, although i haven't been able to do it. It would be helpful if someone can solve this .
 A: If $x>1$
$$x^{12} >x^9 \; ; \;x^4>x$$$$ \implies x^{12} - x^9 + x^4 - x + 1=\color{blue}{(x^{12}-x^9)+(x^4-x)+(1) >0}$$
If $1 >x >0$
$$x^4 > x^9 \; ; \; 1>x $$
$$\implies  x^{12} - x^9 + x^4 - x + 1=\color{blue}{(x^{4}-x^9)+(1-x)+(x^4) >0}$$
If $x \le 0$, each term of the expression inside the root is positive.
Thus, no matter what $x$ is, quantity under the root is positive. Hence, the domain of this function is $\color{red}{x \in \Bbb R}$.
A: There will be other more efficient suggestions given, but here, instead of going for your expression directly, you could consider $$p(x)=x^{12}-x^9+x^4-x=x(x^3-1)(x^8+1)=x(x-1)(x^2+x+1)(x^8+1)$$
You are interested in where this might fall below $-1$ in value. As a twelfth degree polynomial with positive leading coefficient it is ultimately positive when $x$ has large absolute value. The third and fourth factors are always positive and the two real roots of the other factors are $x=0,1$. 
So $p(x)$ therefore falls below zero only when $x\in (0,1)$, and this is the only interval of concern - so $x$ is "small" and positive and the lower powers of $x$ have the greatest absolute value.
Taking advantage of this note that $p(x)+1=(1-x)+(x^4-x^9)+x^{12}$ is the sum of positive terms in this interval.

This was motivated by your comments on the shape of the curve - simply spotting the factors of a related polynomial retrieves that as a possible way forward.
A: The function is in the form $$f(x)=\sqrt{p(x)}$$
and the domain is given by the condition $p(x)\ge 0$ that is
$$x^{12} - x^9 + x^4 - x + 1\ge 0\iff x^{12} + x^4 + 1\ge x^9+x$$
that is always true indeed


*

*for $x<0$ $RHS>0$ and $LHS<0$

*for $0\le x < 1$ we have that $x^4+1\ge x^9+x$

*for $x \ge 1$ we have $x^{12}+x^4\ge x^9+x$


thus the domain is $x\in \mathbb{R}$.
A: $$x^{12}-x^9+x^4-x+1=\frac{1}{2}\bigg[2x^{12}-2x^9+2x^4-2x+2\bigg]$$
$$=\frac{1}{2}\bigg[(x^{12}-2x^9+x^6)+\bigg(x^{12}-x^6+\frac{1}{4}\bigg)+x^4+\bigg(x^4-x^2+\frac{1}{4}\bigg)+(x^2-2x+1)+\frac{1}{2}\bigg]$$
$$=\frac{1}{2}\bigg[(x^6-x^3)^2+\bigg(x^6-\frac{1}{2}\bigg)^2+x^4+\bigg(x^2-\frac{1}{2}\bigg)^2+(x-1)^2+\frac{1}{2}\bigg]>0\forall x\in\mathbb{R}.$$
