Răspuns:
Explicație pas cu pas:
[tex]a)f'(x)=(e^x-x)'=(e^x)'-x'=e^x-1\\b)f'(x)=0\Rightarrow e^x-1=0\\e^x=1\Rightarrow x=0\\\texttt{Dupa ce faci tabelul , se observa ca f este strict crescatoare}\\\texttt{pe }[0,\infty)\texttt{ si strict descrescatoare pe }(-\infty,0).[/tex]
[tex]c)\texttt{Avem ca }\displaystyle\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\infty,\texttt{deci x=0 este punct de minim}\\\texttt{ceea ce inseamna ca }f(x)\geq f(0),\forall x\in\mathbb{R}.\\\texttt{Pentru x}=\dfrac{1}{2},\texttt{inecuatia devine :}\\f\left(\dfrac{1}{2}\right)\geq f(0)\\e^{\frac{1}{2}}-\dfrac{1}{2}\geq 1\\\sqrt{e}\geq 1+\dfrac{1}{2}\\\sqrt{e}\geq \dfrac{3}{2}[/tex]
