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    the curve C has equation f(x)=(x³-2x)e^-x

    a) find f '(x)
    ANSWER: (x³-2x)(-e^-x) + (e^-x)(3x²-2)

    the normal to C at the origin O intersects C at a point P

    b) show that the x-coordinate of P is the solution of the equation 2x²=e^x +4

    if anyone could help with this question that would be great. i have done the first part which is correct, i just don't know how to do part b.
    thanks a lot
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    (Original post by luketennisboy)
    the curve C has equation f(x)=(x³-2x)e^-x

    a) find f '(x)
    ANSWER: (x³-2x)(-e^-x) + (e^-x)(3x²-2)

    the normal to C at the origin O intersects C at a point P

    b) show that the x-coordinate of P is the solution of the equation 2x²=e^x +4

    if anyone could help with this question that would be great. i have done the first part which is correct, i just don't know how to do part b.
    thanks a lot
    grad of tangent at origin is -2 ...........sub x=0 into part (a)
    => grad of norm at origin is 1/2
    eqn of norm at origin is y=1/2x
    intersects C when (x^3-2x)e^(-x)=1/2x
    2(x^2-2)e^-(x)=1
    2x^2-4=e^x
    2x^2=e^x+4
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    how is f` thr first thing correct but why +(e^-x)(3x²-2)
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    (Original post by ABUSHOMAL///M3)
    how is f` thr first thing correct but why +(e^-x)(3x²-2)
    I am not sure what you mean but s/he has used the product rule.
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    plz plz plz plz how is f` the first one ok but why +(e^-x)(3x²-2)
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    how product rule and there is one term
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    ha plz help
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    (Original post by ABUSHOMAL///M3)
    how product rule and there is one term
    f(x)=(x³-2x)e^-x

    Let u = (x³ - 2x) and v = e^-x

    So: du/dx = 3x² - 2 and dv/dx = -e^-x

    So: f '(x) = u(dv/dx) + v(du/dx)
    = (x³ - 2x)(-e^-x) + (3x² - 2)e^-x
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    ahaa thx , by the waay when is ur c3 exam
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    Next Monday (20th).
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    [I]ahaa thx , by the waay when is ur c3 exam]
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    (Original post by evariste)
    grad of tangent at origin is -2 ...........sub x=0 into part (a)
    => grad of norm at origin is 1/2
    eqn of norm at origin is y=1/2x
    intersects C when (x^3-2x)e^(-x)=1/2x
    2(x^2-2)e^-(x)=1
    2x^2-4=e^x
    2x^2=e^x+4
    cheers mate
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    srry but @ what time in GMT i`am a student in JORDAN
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    srry but @ what time in GMT i`am a student in JORDAN
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    srry but @ what time in GMT i`am a student in JORDAN
    srry but @ what time in GMT i`am a student in JORDAN
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    9.00 Bst
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    what is bst
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    what is bst ????
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    British Summer Time
 
 
 
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