Răspuns:
[tex]b)\displaystyle\texttt{Fie }y=mx+n,\ m,n\in\mathbb{R},m\neq 0 ,\texttt{ ecuatia asimptotei oblice}\\m=\lim_{x\to -\infty}\dfrac{f(x)}{x}=\lim_{x\to -\infty}\dfrac{3x+e^x}{x}=3+\lim_{x\to-\infty}\dfrac{e^x}{x}=3-\dfrac{0}{\infty}=3\\n=\lim_{x\to -\infty}(f(x)-3x)=\lim_{x\to -\infty} e^{x}=e^{-\infty}=0\\\texttt{Prin urmare ecuatia asimptotei oblice este }y=3x[/tex]
[tex]c)f(x)\geq 4x+1\\3x+e^x\geq 4x+1\\e^x-x-1\geq 0\\\texttt{Consideram functia }g:\mathbb{R}\rightarrow\mathbb{R},g(x)=e^x-x-1\\g'(x)=e^x-1\\g'(x)=0\Leftrightarrow e^x-1=0\\~~~~~~~~~~~~~~~~~~e^x=1\\~~~~~~~~~~~~~~~~~~~x=0\\~~~~~x|-\infty~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~~~\infty\\g'(x)|~~~-~~~~-~~~0~~~~~~+~~~~~~~~+~~~~\\g(x)|~~~\searrow~~~\searrow~~~~~~~~~\nearrow~~~~\nearrow[/tex]
[tex]\displaystyle\lim_{x\to -\infty}g(x)=\lim_{x\to -\infty}(e^x-x-1)=0+\infty-1=\infty\\\lim_{x\to \infty} g(x)=\lim_{x\to\infty}(e^x-x-1)=\lim_{x\to\infty}e^x\left(1-\dfrac{x}{e^x}-\dfrac{1}{e^x}\right)=\infty\\\texttt{Cum }\lim_{x\to\infty} g(x)=\lim_{x\to -\infty} g(x)=\infty, \texttt{ f este descrescatoare pe }(-\infty,0]}\\\texttt{si crescatoare pe }{[0,\infty), \texttt{iar f este si continua rezulta ca }}\\\texttt{ x=0 este punct de minim global. Prin urmare:}\\g(x)\geq g(0)\forall x\in\mathbb{R}[/tex]
[tex]e^x-x-1\geq 0,\texttt{ ceea ce trebuia demonstrat.} Q.E.D.[/tex]
