Răspuns:
Explicație pas cu pas:
[tex]a)\displaystyle f:(-1,\infty)\rightarrow \mathbb{R}, f(x)=mx-\ln(x+1)\\a)f(x)=x-\ln(x+1)\\\lim_{x\searrow -1}f(x)=-1+\infty=\infty\Rightarrow x=-1\texttt{ asimptota verticala.}\\\lim_{x\to\infty}( x-\ln(x+1))=\lim_{x\to\infty}x\left(1-\dfrac{\ln(x+1)}{x}\right)=\infty\cdot 1=\infty\\!\lim_{x\to\infty} \dfrac{\ln(x+1)}{x}\stackrel{\frac{\infty}{\infty}}{=}\lim_{x\to\infty}\dfrac{1}{x+1}=0[/tex]
[tex]b)f'(x)=1-\dfrac{1}{x+1}=\dfrac{x+1-1}{x+1}=\dfrac{x}{x+1}\\x~~~~~| -1~~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\infty \\f'(x)|~~~-~~-~~-~0~~~+~~~~~~~~~+`~~~~~~~~+\\f(x)~|~~~~~~~~\searrow~~~~~~~~~~~~~~~~~~~~\nearrow[/tex]
[tex]\texttt{f este stric descrescatoare pe }(-1,0]}\texttt{ si strict crescatoare pe}\\{[0,\infty)[/tex]
[tex]\displaystyle c)f'(x)=m-\dfrac{1}{x+1}\\f'(x)=0\Leftrightarrow x=\dfrac{1-m}{m}\\\texttt{Se pune conditia ca }\dfrac{1-m}{m}>-1,\texttt{ de unde rezulta }m>0.\\\texttt{Din tabel se observa ca }\dfrac{1-m}{m}\texttt{ este punct de minim}\\\texttt{(in fine, nu am mai facut tabelul , dar e la fel ca la }\\\texttt{punctul b, doar ca in loc de 0 ai }\dfrac{1-m}{m}),\texttt{ asadar }\\f\left(\dfrac{1-m}{m}\right)\geq 0\\m\cdot \dfrac{1-m}{m}-\ln\left(1+\dfrac{1-m}{m}\right)\geq 0[/tex]
[tex]\displaystyle 1-m-\ln\dfrac{1}{m}\geq 0\\1-m+\ln m\geq 0\\\texttt{Consideram functia }g:(0,\infty)\rightarrow \mathbb{R},g(x)=1-x+\ln x\\g'(x)=-1+\dfrac{1}{x}=\dfrac{1-x}{x}\\g'(x)=0\Leftrightarrow 1-x=0,\texttt{ deci }x=1.\\\texttt{Cum }\lim_{x\searrow 0} g(x)=\lim_{x\to\infty}g(x)=-\infty\texttt{,iar g este crescatoare pe}\\(0,1]\texttt{ si descrescatoare pe }[1,\infty),\texttt{ rezulta ca }g(x)\leq g(1),\\\texttt{deci }g(x)\leq 0\forall x\in(0,\infty)[/tex][tex]\texttt{Egalitatea are loc daca si numai daca }x=1 ,\texttt{ prin urmare}\\\texttt{singura valoare a lui m este 1.}[/tex]
