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For-all-real-values-of-x-solve-the-inequality-1-x-3-x-6-2x-3-5-1-4-




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

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