Question

tot exercitiul 1 de la subiectul 3 daca se poate . multumesc​

Answer 1

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = (x+2)e^{-|x|}\\ \\ |x| ' = \dfrac{x}{|x|}\\ \\ \begin{array}{lcl} f'(x) &=& (x+2)'e^{-|x|}+(x+2)\Big(e^{-|x|}\Big)' \\ \\ &=&e^{-|x|}+(x+2)\Big(-|x|'e^{-|x|}\Big) \\ \\ &=&e^{-|x|}-(x+2)\Big(\dfrac{x}{|x|}\cdot e^{-|x|}\Big)\end{array}\\ \\ \\\underset{x>0}{\lim\limits_{x\to 0}}\dfrac{f-f(0)}{x-0} =\underset{x>0}{\lim\limits_{x\to 0}}\,f'(x) = 1-(0+2)\Big((+1)\cdot e^0\Big) =1-2 = -1$

f nu este derivabila in x = 0 deoarece:

$f':\mathbb{R}\backslash \{0\}\to \mathbb{R},\quad f'(x) =e^{-|x|}-(x+2)\Big(\dfrac{x}{|x|}\cdot e^{-|x|}\Big)$

x ≠ 0 e condiția de existență a lui f'(x).

$f'(x) = e^{-|x|}-(x+2)\Big(\dfrac{x}{|x|}\cdot e^{-|x|}\Big) = 0\\ \\ \Rightarrow e^{-|x|}=(x+2)\cdot \dfrac{xe^{-|x|}}{|x|} \Bigg|:e^{-|x|} \\ \\ \Rightarrow 1 = (x+2)\cdot \dfrac{x}{|x|}\Bigg|\cdot |x| \\\\ \Rightarrow |x| = (x+2)x \\ \\(1)\quad x< 0\Rightarrow -x=(x+2)x \Rightarrow x(x+2+1) = 0 \Rightarrow \\ \Rightarrow x(x+3) = 0 \Rightarrow x = -3\\ \\(2)\quad x>0\Rightarrow x = (x+2)x\Rightarrow x(x+2-1) = 0\Rightarrow x(x+1) = 0 \Rightarrow\\ \Rightarrow x = -1\quad (F)$

=> Punctul de extrem local exte x = -3.