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    Q5 B

    Using product rule I got: xe^-x(2-x)

    So x=2 and it's a maximum

    In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
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    (Original post by Vorsah)
    Q5 B

    Using product rule I got: xe^-x(2-x)

    So x=2 and it's a maximum

    In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
    dy/dx = 0 <=> x(2-x)e^(-x) = 0 <=> x = 0 or 2-x = 0
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    (Original post by Vorsah)
    Q5 B

    Using product rule I got: xe^-x(2-x)

    So x=2 and it's a maximum

    In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
    Note that, since

    e^{-x} \neq 0

    you have

    x(2-x)=0

    which has TWO solutions
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    (Original post by Vorsah)
    Q5 B

    Using product rule I got: xe^-x(2-x)

    So x=2 and it's a maximum

    In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
    Stationary point when dy/dx = 0. When x=0 dy/dx=0 therefore there is a stationary point at x=0
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    Thanks

    But for Q5 A in the attachment I got e^(-2x)(1-2x) using product rule

    So shouldn't it be X=0 or X=0.5

    Because in the answers it only says max at 0.5?
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    (Original post by Vorsah)
    Thanks

    But for Q5 A in the attachment I got e^(-2x)(1-2x) using product rule

    So shouldn't it be X=0 or X=0.5

    Because in the answers it only says max at 0.5?
    No because when x=0 dy/dy=1
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    I need help with Q6 as well.

    I got (3x^2)lnx + x^2

    = x^2(3lnx + 1)

    Don't know what to do from here
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    (Original post by Username_valid)
    No because when x=0 dy/dy=1
    So once you get the2 values of X do u have to put them into the gradient function to see if they = 0

    Eg for Q5 A I got x=0 or x=0.5 , so do I sub in these values into gradient function and see which one = 0 and then use the value that=0 and ignore the one that doesn't = 0?

    When x=0, dy/dx=1

    When x = 0.5 dy/dx=0 so this is the correct value
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    (Original post by Vorsah)
    I need help with Q6 as well.

    I got (3x^2)lnx + x^2

    = x^2(3lnx + 1)

    Don't know what to do from here
    You need to differentiate it and set that equal to 0 to get the stationary points, and then differentiate again to find which is a maximum/minimum.
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    (Original post by The Shanus)
    You need to differentiate it and set that equal to 0 to get the stationary points, and then differentiate again to find which is a maximum/minimum.
    I have differentiated and got (x^2)(3lnx+1)

    So is it x^2 = 0 and

    3lnx+1=0
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    (Original post by Indeterminate)
    Note that, since

    e^{-x} \neq 0

    you have

    x(2-x)=0

    which has TWO solutions
    OP. according to my calculations Indeterminate is correct.
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    (Original post by Vorsah)
    So once you get the2 values of X do u have to put them into the gradient function to see if they = 0

    Eg for Q5 A I got x=0 or x=0.5 , so do I sub in these values into gradient function and see which one = 0 and then use the value that=0 and ignore the one that doesn't = 0?

    When x=0, dy/dx=1

    When x = 0.5 dy/dx=0 so this is the correct value
    You differentiate the function to get dy/dy and equate that to zero. With 5a you get e^-2x(1-2x)=0 but e^-2x cannot equate to zero (ln0 is undefined) so you're left with 1-2x=0 so x=0.5. Find d^2y/dx^2 and sub x=0.5 to determine nature of point.
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    (Original post by Username_valid)
    You differentiate the function to get dy/dy and equate that to zero. With 5a you get e^-2x(1-2x)=0 but e^-2x cannot equate to zero (ln0 is undefined) so you're left with 1-2x=0 so x=0.5. Find d^2y/dx^2 and sub x=0.5 to determine nature of point.
    with 5A

    I get x(e^-x) = x/(e^x)=0 >>> x=0 is this correct ?


    and

    2-x=0
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    (Original post by Vorsah)
    with 5A

    I get x(e^-x) = x/(e^x)=0 >>> x=0 is this correct ?


    and

    2-x=0
    I'm not sure what you did to obtain that, what did you get when you did dy/dx? If you got x/(e^x) then that is incorrect. Use the product rule...
 
 
 
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