Question

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.

Answer 1

`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)= \\= (2r-4)/(e^r) - (r^2-4r)/(e^r)= (2r-4-(r^2-4r))/(e^r)= (2r-4-r^2+4r)/(e^r)= (-r^2+6r-4)/(e^r) \\ \\y'(4)=(-4^2+6* 4-4)/(e^4)= (4)/(e^4) \\ \\2. \\y''(r)=[(-r^2+6r-4)/(e^r)]'=((-r^2+6r-4)'e^(-r)-(-r^2+6r-4)(e^(-r))')/(e^(2r))= \\=((-2r+6)e^(-r)-(-r^2+6r-4)e^(-r)(-1))/(e^(2r))=`

`=((-2r+6)e^(-r)+(-r^2+6r-4)e^(-r))/(e^(2r))=(e^(-r)(-2r+6-r^2+6r-4))/(e^(2r))= \\=(-r^2+4r+2)/(e^(3r)) \\ \\y''(4)=(-4^2+4*4+2)/(e^(3*4))= (2)/(e^(12))`

3. Min and max values

`y'(r)=0`


`(-r^2+6r-4)/(e^r)=0 \\-r^2+6r-4=0 \\r_(1,2)= (-6 \pm √(6^2-4(-1)(-4)) )/(2(-1))=(-6 \pm √(20) )/(-2)= (-6 \pm 2√(5) )/(-2)= 3\mp √(5) \\ a=-1\ \textless \ 0 \Rightarrow parabola \: -r^2+6r-4 \: has \: maximum \\r \in(-\infty,3- √(5)) \cup(3+ √(5) ,\infty) \Rightarrow y'\ \textless \ 0 \Rightarrow y \downarrow \\r \in(3- √(5),3+ √(5)) \Rightarrow y'\ \textgreater \ 0 \Rightarrow y \uparrow \\r_(min)=3- √(5) \\r_(max)=3+√(5)`