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Question Number 132295 by shaker last updated on 13/Feb/21

Commented by mathmax by abdo last updated on 13/Feb/21

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}-\sqrt{5\mathrm{x}-5}\mathrm{f}\mathrm{defined}\mathrm{on}[1,+\mathrm{\infty }[$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}-\frac{5}{2\sqrt{5\mathrm{x}-5}}=\frac{1}{\mathrm{x}}-\frac{\sqrt{5}}{2\sqrt{\mathrm{x}-1}}=\frac{2\sqrt{\mathrm{x}-1}-\sqrt{5}\mathrm{x}}{2\mathrm{x}\sqrt{\mathrm{x}-1}}$$$$=\frac{(2\sqrt{\mathrm{x}-1}-\sqrt{5}\mathrm{x})(2\sqrt{\mathrm{x}-1}+\sqrt{5}\mathrm{x})}{2\mathrm{x}\sqrt{\mathrm{x}-1}(2\sqrt{\mathrm{x}-1}+\sqrt{5}\mathrm{x})}=\frac{4(\mathrm{x}-1)-{5\mathrm{x}}^2}{(...)}=\frac{-{5\mathrm{x}}^2+4\mathrm{x}-4}{(....)}$$$${\mathrm{\Delta }}^{{}^{\prime }}=4-20<0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)<0\Rightarrow \mathrm{f}\mathrm{is}\mathrm{decreazing}\mathrm{on}[1,+\mathrm{\infty }[$$$$\mathrm{f}\left(1\right)=0\mathrm{and}{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\sqrt{5\mathrm{x}}\{\frac{\mathrm{logx}}{\sqrt{5\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}}\}$$$$={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\sqrt{5\mathrm{x}}\{2\frac{\log \left(\sqrt{\mathrm{x}}\right)}{\sqrt{5\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}}\}=-\mathrm{\infty }\Rightarrow \mathrm{\exists }!{\mathrm{x}}_0\in [1,+\mathrm{\infty }[/$$$$\mathrm{f}\left({\mathrm{x}}_0\right)=0\mathrm{and}\mathrm{we}\mathrm{see}\mathrm{that}{\mathrm{x}}_0=1$$

Commented by liki last updated on 13/Feb/21

Commented by liki last updated on 13/Feb/21

$$\mathrm{Help}\mathrm{me}\mathrm{please}$$

Commented by mr W last updated on 13/Feb/21

$$x=1$$

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