![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}}= ](/tpl/images/0217/3020/1baaa.png)
Consider the following function of variable r, for r_0: y(r)=(r^2_ 4.00r)exp(_r).
find the val...
Mathematics, 01.09.2019 08:30 gabrielgutierrez730
Consider the following function of variable r, for r_0: y(r)=(r^2_ 4.00r)exp(_r).
find the value of the first derivative, dy/dr, at r=4.00.
find the value of the second derivative, d^2 y/dr^2, at r=4.00.
find the value of r where y takes its minimum value in the r_0 interval.
find the value of r where y takes its maximum value in the r_0 interval.
Answers: 1
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