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Goten77
 4 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
Goten77
 4 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

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Goten77
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=xe ^{x/8}\] then derivative is this correct? \[y ^{1}=xe ^{x/8}(1/8) + e ^{x/8}1\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1simplifying 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok thats what i got and then i get like 1/8x = 1 then x=1/(1/8) = +8

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1thats correct.... that is the only value of x that makes the 1st derivative = 0...

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.0oh nvm i c y e^o=1 eh i aint thinking

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1if 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
 4 years ago
Best ResponseYou've already chosen the best response.0man i made so many simple mistakes on this last quiz... today i just aint thinking... oh well atleast i got this concept right XD
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