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1-x-5-x-1-4-




Question Number 79560 by jagoll last updated on 26/Jan/20
(√(1+x)) ≤ ((5−x))^(1/(4 ))
$$\sqrt{\mathrm{1}+\mathrm{x}}\:\leqslant\:\sqrt[{\mathrm{4}\:}]{\mathrm{5}−\mathrm{x}} \\ $$
Commented by john santu last updated on 26/Jan/20
(1+x)^2 ≤5−x , x≤5 ∧x≥−1⇒−1≤x≤5  x^2 +2x+1≤5−x  x^2 +3x−4≤0  (x+4)(x−1)≤0  −4≤x≤1 ∧ −1≤x≤5  x ∈ [−1,1]
$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} \leqslant\mathrm{5}−\mathrm{x}\:,\:\mathrm{x}\leqslant\mathrm{5}\:\wedge\mathrm{x}\geqslant−\mathrm{1}\Rightarrow−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{5} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\leqslant\mathrm{5}−\mathrm{x} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{4}\leqslant\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}−\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$−\mathrm{4}\leqslant\mathrm{x}\leqslant\mathrm{1}\:\wedge\:−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{5} \\ $$$$\mathrm{x}\:\in\:\left[−\mathrm{1},\mathrm{1}\right] \\ $$

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