Răspuns:
Explicație pas cu pas:
[tex]a)\displaystyle\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0\Rightarrow y=0~\texttt{asimptota orizontala.}\\\texttt{Pentru asimptota verticala se calculeaza limita in punctele de}\\\texttt{de discontinuitate.}\\\lim_{x\to -1}f(x)=-\infty\\\lim_{x\to 0}f(x)=+\infty\\\texttt{Prin urmare, x=-1 si x=0 sunt asimptote verticale.}\\\texttt{Deoarece functia admite asimptota orizontala, nu mai are rost}\\\texttt{sa calculam asimptota oblica.}[/tex]
[tex]b)\texttt{Functia mai poate fi scrisa sub forma:}\\f(x)=\dfrac{(x+1)^2-x^2}{x^2(x+1)^2}=\dfrac{1}{x^2}-\dfrac{1}{(x+1)^2}\\\\f'(x)=-\dfrac{2}{x^3}+\dfrac{2}{(x+1)^3}=2\cdot \dfrac{(x+1)^3-x^3}{x^3(x+1)^3}=\\2\cdot\dfrac{(x+1-x)((x+1)^2+x(x+1)+x^2)}{x^3(x+1)^3}=\\2\cdot\dfrac{x^2+2x+1+x^2+x+x^2}{x^3(x+1)^3}=2\cdot \dfrac{3x^2+3x+1}{x^3(x+1)^3}\\f'(x)=0\Leftrightarrow 3x^2+3x+1=0\\~~~~~~~~~~~~~~~~~\Delta=9-4\cdot3=-3<0,\texttt{deci ecuatia nu are solutii reale}[/tex]
[tex]\texttt{Prin urmare, f nu admite puncte de extrem local.}[/tex]
