y=xe^(-x/8)
y1=x(e^(-x/8))(-1/8) + e^(-x/8)(1)
critical points
ima type this below along w/ the question

Question

y=xe^(-x/8)
y1=x(e^(-x/8))(-1/8) + e^(-x/8)(1)
critical points
ima type this below along w/ the question

goten77 · Answer

$$y=xe ^{-x/8}$$  then derivative is this correct?
$$y ^{1}=xe ^{-x/8}(-1/8) + e ^{-x/8}1$$

campbell_st · Answer

simplifying your 1st derivative you will have 

$$\frac{dy}{dx} = e^{-\frac{x}{8}}(-\frac{1}{8} x + 1)$$

so you need to solve 

$$-\frac{1}{8} x + 1 = 0$$

goten77 · Answer

ok thats what i got and then i get like
-1/8x = -1   then 
x=-1/(-1/8) = +8

campbell_st · Answer

thats correct.... that is the only value of x that makes the 1st derivative = 0...

goten77 · Answer

oh nvm i c y e^o=1 eh i aint thinking

campbell_st · Answer

if you have x = 0 then 

$$dy/dx = e^{-\frac{0}{8}}(-\frac{1}{8}\times 0 + 1) = 1$$

because  

$$e^{\frac{0}{8}} = 1$$

any number to a power of zero is 1

so only 1 critical point...

goten77 · Answer

man i made so many simple mistakes on this last quiz... today i just aint thinking... oh well atleast i got this concept right XD