[tex]f(x) = \dfrac{e^{|\ln x|}}{x+1} \\ \\ f'(x) = \Bigg(\dfrac{e^{\sqrt{\ln^2 x}}}{x+1}\Bigg)' = \dfrac{\dfrac{2\ln x}{2x|\ln x|}\cdot e^{|\ln x|}(x+1)-e^{|\ln x|}}{(x+1)^2} = \\ \\\\ = \dfrac{(x+1)\ln x\cdot e^{|\ln x|}-x|\ln x|e^{|\ln x|}}{x|\ln x|(x+1)^2} \\ \\ f'(x) = 0[/tex]
[tex]\boxed{1}\quad x\geq 1 \Rightarrow |\ln x| =\ln x \Rightarrow \\ \\ \Rightarrow (x+1)\ln x\cdot e^{|\ln x|}-x\ln x \cdot e^{|\ln x|} = 0 \Rightarrow \\ \\ \Rightarrow \ln x\cdot e^{|\ln x|} = 0 \Rightarrow x = 1[/tex]
x = 1 este punct unghiular, deoarece derivata nu are sens în acel punct.
[tex]\boxed{2}\quad x<1\Rightarrow |\ln x| =-\ln x \Rightarrow \\ \\ \Rightarrow (x+1)\ln x\cdot e^{|\ln x|}+x\ln x\cdot e^{|\ln x|} = 0 \Rightarrow \\ \\\Rightarrow\ln x\cdot e^{|\ln x|}(x+1+x) = 0 \Rightarrow x \in \Big\{-\dfrac{1}{2},1\Big\} \quad (Fals) [/tex]
[tex]\Rightarrow \text{Singurul punct de extrem local este }\Big(1,f(1)\Big) \text{ (punct unghiular)}[/tex]
Răspuns corect d).
