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Goten77

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

  • one year ago
  • one year ago

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  1. Goten77
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    \[y=xe ^{-x/8}\] then derivative is this correct? \[y ^{1}=xe ^{-x/8}(-1/8) + e ^{-x/8}1\]

    • one year ago
  2. campbell_st
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    simplifying your 1st derivative you will have \[\frac{dy}{dx} = e^{-\frac{x}{8}}(-\frac{1}{8} x + 1)\] so you need to solve \[-\frac{1}{8} x + 1 = 0\]

    • one year ago
  3. Goten77
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    ok thats what i got and then i get like -1/8x = -1 then x=-1/(-1/8) = +8

    • one year ago
  4. campbell_st
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    thats correct.... that is the only value of x that makes the 1st derivative = 0...

    • one year ago
  5. Goten77
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    oh nvm i c y e^o=1 eh i aint thinking

    • one year ago
  6. campbell_st
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    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...

    • one year ago
  7. Goten77
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    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

    • one year ago
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