Stationary points core 3

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    stauionary points on the curve y = x^2 e ^ -x
    y' = xe^-x

    so x=0 and x=2

    y"= 2x(e^-x) + - e^-x(2)

    at 0,0 max point

    at 2,0.541 min point

    i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
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    (Original post by Custardcream000)
    stauionary points on the curve y = x^2 e ^ -x
    y' = xe^-x

    so x=0 and x=2

    y"= 2x(e^-x) + - e^-x(2)

    at 0,0 max point

    at 2,0.541 min point

    i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
    What's your question?
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    (Original post by Custardcream000)
    stauionary points on the curve y = x^2 e ^ -x
    y' = xe^-x

    so x=0 and x=2

    y"= 2x(e^-x) + - e^-x(2)

    at 0,0 max point

    at 2,0.541 min point

    i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
    Don't use decimals. Leave it in exact form.
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    (Original post by Zacken)
    What's your question?
    find the stationary points on the curve...
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    (Original post by RDKGames)
    Don't use decimals. Leave it in exact form.
    x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
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    (Original post by Custardcream000)
    find the stationary points on the curve...
    Your first derivative is wrong.
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    (Original post by Custardcream000)
    x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
    y = \frac{4}{e^2} is the exact form.... you've literally written it out...

    ...
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    (Original post by Custardcream000)
    x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
    ..........


    Yes there is. You literally wrote it out right after the equals sign on y.
 
 
 
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