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Question Number 56356 by Tawa1 last updated on 15/Mar/19

$$\mathrm{Find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{F}\left(\mathrm{x}\right)={\log }_{\mathrm{e}}\mathrm{x}-\mathrm{x}\mathrm{for}\mathrm{x}>0.$$ $$\mathrm{Hence}\mathrm{show}\mathrm{that}:{\log }_{\mathrm{e}}\mathrm{x}⩽\mathrm{x}-1.\mathrm{For}\mathrm{all}\mathrm{x}>0$$

Commented byTawa1 last updated on 15/Mar/19

Answered by kaivan.ahmadi last updated on 15/Mar/19

$$f\left(x\right)=lnx-x\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}-1=0\Rightarrow x=1$$ $$ifx<1\Rightarrow \frac{1}{x}>1\Rightarrow \frac{1}{x}-1>0\Rightarrow f^{\prime }\left(x\right)>0$$ $$somillarlyifx>1\Rightarrow f^{\prime }\left(x\right)<0$$ $$\Rightarrow f\left(1\right)=-1ismaximum\Rightarrow$$ $$f\left(x\right)⩽-1\Rightarrow lnx-x⩽-1\Rightarrow$$ $$lnx⩽x-1$$ $$$$ $$$$

Commented byTawa1 last updated on 15/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19

$$\frac{df\left(x\right)}{dx}=\frac{1}{x}-1$$ $$\frac{d^{2}f}{{dx}^2}=\frac{-1}{x^2}$$ $$formin/max\frac{df}{dx}=0$$ $$so\frac{1}{x}-1=0$$ $$x=1$$ $$atx=1\frac{d^{2}f}{{dx}^2}=\frac{-1}{1^2}<0$$ $$sonominimumvalue...$$ $$formin\frac{d^{2}f}{{dx}^2}>0andformax\frac{d^{2}f}{{dx}^2}<0$$ $$soatx=1f\left(x\right)=\mid lnx-x{\mid }_{x=1}\to -1$$ $$value=-1$$

Commented bytanmay.chaudhury50@gmail.com last updated on 15/Mar/19

Commented bytanmay.chaudhury50@gmail.com last updated on 15/Mar/19

$$fromgraphoflnx-xitisclearthatf\left(x\right)hssmaxvalueatx=1andthatvalueis-1$$

Commented byTawa1 last updated on 15/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{effort}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19

$$if\frac{dg}{dx}<\frac{dh}{dx}$$ $$theng\left(x\right)<h\left(x\right)$$ $$now\frac{d}{dx}\left(lnx\right)=\frac{1}{x}$$ $$\frac{d}{dx}(x-1)=1$$ $$now\frac{1}{x}-1$$ $$\frac{1-x}{x}<0whenx>1$$ $$solnx<(x-1)whenx>1$$ $$atx=1i,eln\left(1\right)=0$$ $$(1-1)=0$$ $${\left(lnx\right)}_{x=1}={(x-1)}_{x=1}$$ $$when1>x>0$$ $$lnx=-ve(x-1)=-ve$$ $$and\mid lnx\mid >\mid (x-1)\mid$$ $$solnx<(x-1)when1>x>0$$ $$solnx<(x-1)whenx>0$$ $$$$

Commented bytanmay.chaudhury50@gmail.com last updated on 15/Mar/19

Commented byTawa1 last updated on 15/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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