Respuesta :
[tex]y(r)=(r^2-4r)e^{-r},r \geq 0
\\
\\1.
\\y'(r)=[(r^2-4r)e^{-r}]'=(r^2-4r)'e^{-r}+(r^2-4r)(e^{-r})'=
\\=(2r-4)e^{-r}+(r^2-4r)e^{-r}(-1)=(2r-4)e^{-r}-(r^2-4r)e^{-r}=
\\= \frac{2r-4}{e^r} - \frac{r^2-4r}{e^r}= \frac{2r-4-(r^2-4r)}{e^r}= \frac{2r-4-r^2+4r}{e^r}= \frac{-r^2+6r-4}{e^r}
\\
\\y'(4)=\frac{-4^2+6\times 4-4}{e^4}= \frac{4}{e^4}
\\
\\2.
\\y''(r)=[\frac{-r^2+6r-4}{e^r}]'=\frac{(-r^2+6r-4)'e^{-r}-(-r^2+6r-4)(e^{-r})'}{e^{2r}}=
\\=\frac{(-2r+6)e^{-r}-(-r^2+6r-4)e^{-r}(-1)}{e^{2r}}=
[/tex]
[tex]=\frac{(-2r+6)e^{-r}+(-r^2+6r-4)e^{-r}}{e^{2r}}=\frac{e^{-r}(-2r+6-r^2+6r-4)}{e^{2r}}= \\=\frac{-r^2+4r+2}{e^{3r}} \\ \\y''(4)=\frac{-4^2+4\times4+2}{e^{3\times4}}= \frac{2}{e^{12}} [/tex]
3. Min and max values
[tex]y'(r)=0[/tex]
[tex] \frac{-r^2+6r-4}{e^r}=0 \\-r^2+6r-4=0 \\r_{1,2}= \frac{-6 \pm \sqrt{6^2-4(-1)(-4)} }{2(-1)}=\frac{-6 \pm \sqrt{20} }{-2}= \frac{-6 \pm 2\sqrt{5} }{-2}= 3\mp \sqrt{5} \\ a=-1\ \textless \ 0 \Rightarrow parabola \: -r^2+6r-4 \: has \: maximum \\r \in(-\infty,3- \sqrt{5}) \cup(3+ \sqrt{5} ,\infty) \Rightarrow y'\ \textless \ 0 \Rightarrow y \downarrow \\r \in(3- \sqrt{5},3+ \sqrt{5}) \Rightarrow y'\ \textgreater \ 0 \Rightarrow y \uparrow \\r_{min}=3- \sqrt{5} \\r_{max}=3+\sqrt{5}[/tex]
[tex]=\frac{(-2r+6)e^{-r}+(-r^2+6r-4)e^{-r}}{e^{2r}}=\frac{e^{-r}(-2r+6-r^2+6r-4)}{e^{2r}}= \\=\frac{-r^2+4r+2}{e^{3r}} \\ \\y''(4)=\frac{-4^2+4\times4+2}{e^{3\times4}}= \frac{2}{e^{12}} [/tex]
3. Min and max values
[tex]y'(r)=0[/tex]
[tex] \frac{-r^2+6r-4}{e^r}=0 \\-r^2+6r-4=0 \\r_{1,2}= \frac{-6 \pm \sqrt{6^2-4(-1)(-4)} }{2(-1)}=\frac{-6 \pm \sqrt{20} }{-2}= \frac{-6 \pm 2\sqrt{5} }{-2}= 3\mp \sqrt{5} \\ a=-1\ \textless \ 0 \Rightarrow parabola \: -r^2+6r-4 \: has \: maximum \\r \in(-\infty,3- \sqrt{5}) \cup(3+ \sqrt{5} ,\infty) \Rightarrow y'\ \textless \ 0 \Rightarrow y \downarrow \\r \in(3- \sqrt{5},3+ \sqrt{5}) \Rightarrow y'\ \textgreater \ 0 \Rightarrow y \uparrow \\r_{min}=3- \sqrt{5} \\r_{max}=3+\sqrt{5}[/tex]
