# Domain of the Function Square Root of 12th Degree Polynomial

Find the Domain of $$f(x)=\frac{1}{\sqrt{x^{12}-x^9+x^4-x+1}}$$

My Try: The Domain is given by

$$x^{12}-x^9+x^4-x+1 \gt 0$$ $\implies$

$$x(x-1)(x^2+x+1)(x^8+1)+1 \gt 0$$

-
I am sure that the Domain includes the set $(-\infty \: 0]U[1 \: \infty)$ because if $x(x-1)\gt0$ then $x(x-1)(x^2+x+1)(x^8+1)\gt 0$ – Ekaveera Kumar Sharma May 22 '14 at 12:11
Isn't the domain $(-\infty,\infty)$ ? – lsp May 22 '14 at 12:15
@lsp It is; but how dow we show it...? – gebruiker May 22 '14 at 12:18
This function seems to be defined everywhere ... but I am not able to prove it. Sorry ! – Claude Leibovici May 22 '14 at 12:19

You need to show $x^{12}-x^9+x^4-x+1 \gt 0$ holds for all $x \in \mathbb R$. This is easy if you do not over-brain it.

First note that if $x< 0$, every term is positive, and hence the LHS positive. So we need to worry only about $x \ge 0$.

Here we can get some help from considering two cases. For $0 \le x < 1$, note the LHS can be written as $x^{12} + (x^4-x^9) + (1-x) > 0$ which is obvious. For $x \ge 1$, note the LHS can be written as $(x^{12}-x^9) + (x^4-x) + 1 > 0$.

-
+1 for simpleness. I mean who thinks about this when we know how to differentiate... – evil999man May 22 '14 at 14:41

If $x$ is out of the interval $[0,1]$ all the facotrs in $$x(x-1)(x^2+x+1)(x^8+1)+1$$ are positive so we have $$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$ But for $x\in [0,1],$ $$x(x-1)\geq -\frac{1}{4}$$ So $$x(x-1)(x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot (x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot\frac{3}{4}.1+1=\frac{13}{16}$$ i.e., for all $x\in \mathbb R$ $$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$

-
Where does $.1$ come from in the fourth equation after $\frac34$? – Ruslan May 22 '14 at 14:10