Question Number 128896 by ZiYangLee last updated on 11/Jan/21
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{even}\:\mathrm{obtaining}\:\mathrm{the}\:\mathrm{zero}\left(\mathrm{s}\right), \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{zero}. \\ $$$${f}\left({t}\right)=\left(\mathrm{1}+\sqrt{\mathrm{2}}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{t}^{\mathrm{2}} \left({t}+\sqrt{\mathrm{2}}\right) \\ $$
Answered by Olaf last updated on 11/Jan/21
$$ \\ $$$${f}\left({t}\right)\:=\:\left(\mathrm{1}+\sqrt{\mathrm{2}}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{t}^{\mathrm{2}} \left({t}+\sqrt{\mathrm{2}}\right) \\ $$$${f}'\left({t}\right)\:=\:\sqrt{\mathrm{2}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)−\mathrm{2}{t}\left(\mathrm{1}+\sqrt{\mathrm{2}}{t}\right)+\mathrm{2}{t}\left({t}+\sqrt{\mathrm{2}}\right)+{t}^{\mathrm{2}} \\ $$$${f}'\left({t}\right)\:=\:\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){t}^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){t}+\sqrt{\mathrm{2}} \\ $$$$\Delta\:=\:\mathrm{2}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{3}×\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{2}}\:=\:\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{12}\:<\:\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{sign}\:\mathrm{of}\:{f}'\:\mathrm{is}\:\mathrm{the}\:\mathrm{sign}\:\mathrm{of}\:\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:>\:\mathrm{0} \\ $$$$\Rightarrow\:{f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{function} \\ $$$$\mathrm{and}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{t}^{\mathrm{3}} \:=\:−\infty \\ $$$$\mathrm{and}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)\:=\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{t}^{\mathrm{3}} \:=\:+\infty \\ $$$$\Rightarrow\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}. \\ $$