[tex] \displaystyle Calculam~derivata:~f'(x)=\frac{x+1}{\sqrt{x^2+2x+3}}-1.\\ \\ Observam~ca~f'(x)= \frac{x+1}{\sqrt{(x+1)^2+2}}-1<1-1=0.\\ \\ Deci~f~este~strict~descrescatoare. \\ \\ Inseamna~ca~pentru~a \le b,~avem~f(a) \ge f(b),~si~desigur \\ \\ pentru~a \ge b,~avem~f(a) \le f(b).\\ \\ Iar~a<b~(respectiv~a>b),~vom~avea~f(a)>f(b) \\ \\ (respectiv~f(a)<f(b)).\\ \\ [/tex]
[tex] \displaystyle Noi~trebuie~sa~gasim~varianta~in~care~are~loc~una~dintre \\ \\ situatiile~de~mai~sus.~Din~fericire~varianta~d)~iese~in~evidenta \\ \\ pentru~ca~ne~aduce~aminte~de~o~inegalitate~foarte~importanta \\ \\ facuta~in~clasa~a~11-a:~e^x \ge x+1~\forall~x \in \mathbb{R}. \\ \\ Deci~avem~e^x \ge x+1~si~f(e^x) \le f(x+1),~adevarat! \\ \\ (a \ge b \Rightarrow f(a) \le f(b)) [/tex]
[tex] \displaystyle Raspuns:~d). \\ \\ Voi~adauga~in~comentarii~niste~informatii. [/tex]
