от ammornil » 08 Дек 2012, 21:54
[tex]y=x^2.e^{-x} \hspace{12} \Leftrightarrow \hspace{12} y=\frac{x^2}{e^x} \\
\cyr{DM}: \hspace{12} \forall x \in R \Rightarrow x \in (-\infty; +\infty)[/tex]
***
[tex]y'=\frac{2.x.e^{x}-x^2.e^{x}}{e^{2.x}}=\frac{\cancel{e^x}.(2.x-x^2)}{(e^{x})\cancel{^{2}}}=\frac{-x.(x-2)}{e^x}[/tex]
[tex]y'=0 \hspace{12} \begin{array}{|c}x_1=0 \\ x_2=2 \end{array}[/tex]
***
[tex]: x \in (-\infty;0) \hspace{8} y'<0 \hspace{8} \Rightarrow y-\cyr{monotonno namalyavashcha} \\
: x \in (0;2) \hspace{8} y'>0 \hspace{8} \Rightarrow y-\cyr{monotonno rastyashcha} \\
: x \in (2;+\infty) \hspace{8} y'<0 \hspace{8} \Rightarrow y-\cyr{monotonno namalyavashcha} \\
y(0)=0 -\cyr{lokalen minimum} \\
y(2)=\frac{4}{e^2} -\cyr{lokalen maksimum}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]