Răspuns:
Explicație pas cu pas:
[tex]c)~ f~'(x)=\dfrac{\sqrt{x}(lnx+1) }{x}, ~a~ fost~aratat~in~subpunctul~a)\\Aflam~punctele~de~extrem,~~f~'(x)=0,~deoarece~x>0,~rezulta~lnx+1=0,~deci~~lnx=-1,~~x=e^{-1}=\frac{1}{e} .\\Pentru~x<\frac{1}{e} ,~fie~x=\frac{1}{e^{2}},~atunci~f~'(\frac{1}{e^{2}})=\dfrac{\sqrt{\frac{1}{e^{2}}}(ln\frac{1}{e^{2}}+1) }{\frac{1}{e^{2}}} =\dfrac{\sqrt{\frac{1}{e^{2}}}(-2+1) }{\frac{1}{e^{2}}} <0\\Pentru~x=1>\frac{1}{e} ,~atunci,~f~'(1)=\frac{\sqrt{1}(ln1+1) }{1}>0[/tex]
deci x=1/e este punct de minim. Atunci f(x)≥f(1/e), dar f(1/e)=-4/√e.
[tex]Deci~f(x)\geq -\frac{4}{\sqrt{e} }.~~atunci~\sqrt{e}*f(x)\geq \sqrt{e}*(-\frac{4}{\sqrt{e} }),~~atunci~ \sqrt{e}*f(x)\geq-4,~~atunci~\sqrt{e}*f(x)+4\geq -4+4,~~deci~\sqrt{e}*f(x)+4\geq 0.[/tex]
