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Question Number 67983 by peter frank last updated on 03/Sep/19

Commented by Abdo msup. last updated on 03/Sep/19

$$b)letf\left(x\right)={\left(\frac{1}{x}\right)}^x\Rightarrow f\left(x\right)=x^{-x}=e^{-xln\left(x\right)}$$$$fis?definedon]0,+\mathrm{\infty }[$$$${lim}_{x\to 0andx>0}f\left(x\right)=e^{0=1}$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)={lim}_{x\to +\mathrm{\infty }}{e}^{-xlnx}=0$$$$f^{{}^{\prime }}\left(x\right)=(-lnx-1)e^{-xlnx}=-(lnx+1)f\left(x\right)$$$$f^{{}^{\prime }}\left(x\right)=0\iff lnx+1=0\iff lnx=-1\iff x=e^{-1}$$$$\left(seethstf\right)x>0)$$$$f\left(e^{-1}\right)=e^{-e^{-1}ln\left(e^{-1}\right)}=e^{\frac{1}{e}}$$$$f^{{}^{\prime }}\left(x\right)>0\iff lnx+1<0\iff lnx<-1\iff x<e^{-1}$$$$varistionoff$$$$x0e^{-1}+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)+0-$$$$f\left(x\right)1incr{e}^{\frac{1}{e}}dcres0$$$$\Rightarrow maxf{\left(x\right)}_{x>0}=f\left(e^{-1}\right)=e^{\frac{1}{e}}$$

Answered by MJS last updated on 03/Sep/19

$$a)\mathrm{geometrically}:$$$$\mathrm{any}\mathrm{rectangular}\mathrm{triangle}\mathrm{is}\mathrm{one}\mathrm{half}\mathrm{of}\mathrm{a}$$$$\mathrm{rectangle},\mathrm{the}\mathrm{hypothenuse}\mathrm{of}\mathrm{the}\mathrm{triangle}$$$$\mathrm{is}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{diagonals}\mathrm{of}\mathrm{the}\mathrm{rectangle},\mathrm{the}$$$$\mathrm{diagonals}\mathrm{bisect}\mathrm{each}\mathrm{other},\mathrm{the}{\mathrm{rectangle}}^{\prime }\mathrm{s}$$$$\mathrm{circumcircle}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{the}{\mathrm{triangle}}^{\prime }\mathrm{s},$$$$\mathrm{done}$$

Commented by peter frank last updated on 03/Sep/19

$$thanks$$

Answered by MJS last updated on 03/Sep/19

$$b)$$$$f\left(x\right)=x^{-x}$$$$f^{\prime }\left(x\right)=-(1+\ln x)x^{-x}$$$$f^{\prime }\left(x\right)=0\Rightarrow \ln x=-1\Rightarrow x={\mathrm{e}}^{-1}$$$$f^{″}\left(x\right)=(x{(1+\ln x)}^2-1)x^{-(x+1)}$$$$f^{″}\left({\mathrm{e}}^{-1}\right)=-{\mathrm{e}}^{1+\frac{1}{\mathrm{e}}}<0\Rightarrow$$$$\Rightarrow \mathrm{maximum}\mathrm{at}x={\mathrm{e}}^{-1};f\left({\mathrm{e}}^{-1}\right)={\mathrm{e}}^{\frac{1}{\mathrm{e}}}$$

Commented by peter frank last updated on 03/Sep/19

$$thankyou$$

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