Question
Let $$f\left( x \right) = {\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)^{ - \frac{1}{2}}}.$$ The domain of the function is :
A.
$$\left( {1,\, + \infty } \right)$$
B.
$$\left( { - \infty ,\, - 1} \right)$$
C.
$$\left( { - 1,\,1} \right)$$
D.
$$\left( { - \infty ,\, + \infty } \right)$$
Answer :
$$\left( { - \infty ,\, + \infty } \right)$$
Solution :
$$f\left( x \right)$$ is real for those values of $$x$$ for which
$$\eqalign{
& {x^{12}} - {x^9} + {x^4} - x + 1 > 0 \cr
& {\text{or }}{x^4}\left( {{x^8} + 1} \right) - x\left( {{x^8} + 1} \right) + 1 > 0 \cr
& {\text{or }}\left( {{x^8} + 1} \right)x\left( {{x^3} - 1} \right) + 1 > 0 \cr} $$
If $$x \geqslant 1$$ or $$x \leqslant - 1$$ then the above expression is obviously positive.
If $$ - 1 < x \leqslant 0,$$ the above holds.
If $$0 < x < 1,\,{x^{12}} - x\left( {{x^8} + 1} \right) + \left( {{x^4} + 1} \right) > 0$$ because $${x^4} + 1 > {x^8} + 1$$ and so $${x^4} + 1 > x\left( {{x^8} + 1} \right)$$