Question Number 41095 by Necxx last updated on 02/Aug/18
$${For}\:{all}\:{real}\:{values}\:{of}\:{x}\:{solve}\:{the} \\ $$$${inequality} \\ $$$$\:\mid\frac{\mathrm{1}−{x}^{\mathrm{3}} }{{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}}\mid\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 02/Aug/18
$${let}\:{x}^{\mathrm{3}} −\mathrm{1}={t} \\ $$$$\Rightarrow\:\:\mid\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{4}}\mid\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${or}\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\leqslant\:\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{4}}\:\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${or}\:\:\:\:{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{4}\:\geqslant\:\mathrm{0}\:\:\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\:{t}\in\left(−\infty,\infty\right)\:\: \\ $$$${the}\:{same}\:{for}\:{x}. \\ $$