## Goten77 3 years ago 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

1. Goten77

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

2. campbell_st

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

3. Goten77

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

4. campbell_st

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

5. Goten77

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

6. campbell_st

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...

7. Goten77

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