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Question Number 96586 by  M±th+et+s last updated on 02/Jun/20
find the minimum value of                        f(x)=x^x   for x∈R^+
$${find}\:{the}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x}^{{x}} \\ $$$${for}\:{x}\in\mathbb{R}^{+} \\ $$
Answered by 1549442205 last updated on 02/Jun/20
y=x^x ⇒lny=xlnx⇒((y′)/y)=lnx+1  ⇒y′=x^x (lnx+1)=0⇔lnx=−1  ⇔x=e^(−1) .y′′=(x^x /x)+(lnx+1).x^x (lnx+1)  =x^x ((1/x)+(lnx+1)^2 ).y”(e^(−1) )=  e^(−e^(−1) ) .e=e^(1−e^(−1) ) >0.Hence  miny=minf(x)=f(e^(−1) )=e^(−e^(−1) )
$$\mathrm{y}=\mathrm{x}^{\mathrm{x}} \Rightarrow\mathrm{lny}=\mathrm{xlnx}\Rightarrow\frac{\mathrm{y}'}{\mathrm{y}}=\mathrm{lnx}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}'=\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right)=\mathrm{0}\Leftrightarrow\mathrm{lnx}=−\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{x}=\mathrm{e}^{−\mathrm{1}} .\mathrm{y}''=\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}}+\left(\mathrm{lnx}+\mathrm{1}\right).\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right) \\ $$$$=\mathrm{x}^{\mathrm{x}} \left(\frac{\mathrm{1}}{\mathrm{x}}+\left(\mathrm{lnx}+\mathrm{1}\right)^{\mathrm{2}} \right).\mathrm{y}''\left(\mathrm{e}^{−\mathrm{1}} \right)= \\ $$$$\mathrm{e}^{−\mathrm{e}^{−\mathrm{1}} } .\mathrm{e}=\mathrm{e}^{\mathrm{1}−\mathrm{e}^{−\mathrm{1}} } >\mathrm{0}.\mathrm{Hence} \\ $$$$\mathrm{miny}=\mathrm{minf}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{e}^{−\mathrm{1}} \right)=\mathrm{e}^{−\mathrm{e}^{−\mathrm{1}} } \\ $$
Commented by  M±th+et+s last updated on 03/Jun/20
nice work
$${nice}\:{work} \\ $$

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