Question

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

• A. $(-\infty ,-1)$
• B. $(1,\infty )$
• C. $(-1,1)$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is D $(-\infty ,\infty )$

$f\left(x\right)$ is defined for $x^{12}-x^9+x^4-x+1>0$
$\Rightarrow x^4(x^8+1)-x(x^8+1)+1>0$

$\Rightarrow (x^8+1)x(x^3-1)+1>0$
If $x\ge 1$ or $x\le -1$, then the above expression is positive.
If $-1<x\le 0$, the above inequality still holds.
If $0<x<1$, then $x^{12}-x(x^8+1)+(x^4+1)>0$
$[\because x^4+1>x^8+1$ and so $x^4+1>x(x^8+1)]$.
The domain of $f=(-\infty ,\infty )$