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    I need a step-by-step guide on answering Question 1(a) and (b) as follows.
    Help would be much appreciated.



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    I attempted (a) and I know I've screwed up somewhere here.
    I used the Product Rule for question (a).

    y = e^-4x (x^2 + 2x - 2)

    u = e^-4x
    v = (x^2 + 2x - 2)

    du/dx = -4e^-4x
    Using the Chain Rule I got dv/dx = 2x + 2

    Applying the Product Rule [dy/dx = u(dv/dx) + v(du/dx)]
    I went onto get something different to the answer they wanted me to get, even through factorising. I've messed something up.

    I would also like help on doing the second one. It's a five mark question and I don't know where to start as I can't recall C1 that well.

    Thanks in advance!
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    Well show what you got using the Product Rule (you didn't need chain rule btw).
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    You've got everything correct that you've written there, so your mistake will be after you've used the product rule. For the second part, the stationary points occur when dy/dx=0, so let dy/dx=0 and solve the equation to get the x values of the coordinates, then substitute it into the original equation to get the y values.
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    Yes I thought it was after the Product Rule too, writing up what I got is kind of long.

    Could someone just complete the remainder of question (a) for me?
    That way I'll know where I did actually mess it up
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    Oh and thanks a bunch for the replies!
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    \frac{dy}{dx}= e^{-4x}(2x+2) -4e^{-4x}(x^2+2x-2)

    Is what you have obtained using the product rule. If you take out a factor of e^{-4x} then simplify the term in the bracket and take out another factor, then you should be able to obtain the answer in the required form.
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y\ =\ (e^-4x )(x^2 +2x-2)



u= e^-4x

\dfrac{du}{dx}\ = -4e^-4x



v= x^2 +2x-2

\dfrac{dv}{dx}\ = 2x+2



Product Rule = u dv/dx + v du/dx



= (e^-4x)(2x+2) - 4e^-4x (x^2 +2x-2)



= 2(e^-4x)(x+1) - 4e^-4x (x^2 +2x-2)



= 2(e^-4x)(x+1) - 2(e^-4x) 2(x^2 +2x-2)



= 2e^-4x ((x+1) - 2(x^2 +2x-2))



= 2e^-4x (5-3x-2x^2 )

    For part B, just equal \frac{dy}{dx} to 0
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    Thanks again.

    I'm however still a bit stuck on the stationary points part though, I know I have to equal my dy/dx to 0 but what do I do after that to determine the co-ordinates?
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    (Original post by hass147)
    Thanks again.

    I'm however still a bit stuck on the stationary points part though, I know I have to equal my dy/dx to 0 but what do I do after that to determine the co-ordinates?
    Solve the equation dy/dx=0 to find the x values of the coordinates, then substitute the x values into the original equation to find the y values of the coordinates.
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    (Original post by Phil_Waite)
    Solve the equation dy/dx=0 to find the x values of the coordinates, then substitute the x values into the original equation to find the y values of the coordinates.
    2e^-4x (5 - 3x - 2x^2) = 0

    Divided both sides by 2e^-4x

    (5 - 3x - 2x^2) = 0

    -2x^2 - 3x + 5 = 0

    Quadratic Formula:


    -(-3) +/- root^[ (-3)^2 -4(-2)(5)]
    _______________________
    2(-2)

    x = -5/2
    x = 1


    Is that correct so far?
    Plugging them back into the equation confused me because I don't know if I'm supposed to substitute them into e^-4x's "x" or not and despite that, my answers come out as really small decimals or very high numbers. So I think something's amiss.
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    (Original post by hass147)
    2e^-4x (5 - 3x - 2x^2) = 0

    Divided both sides by 2e^-4x

    (5 - 3x - 2x^2) = 0

    -2x^2 - 3x + 5 = 0

    Quadratic Formula:


    -(-3) +/- root^[ (-3)^2 -4(-2)(5)]
    _______________________
    2(-2)

    x = -5/2
    x = 1


    Is that correct so far?
    Plugging them back into the equation confused me because I don't know if I'm supposed to substitute them into e^-4x's "x" or not and despite that, my answers come out as really small decimals or very high numbers. So I think something's amiss.
    Yes that's correct, but you didn't need to use the quadratic formula because it easily factorises. Substitute them into all of the Xs in the original equation, for example:

    y=e^{-4(1)}((1)^2+2(1)-2)

    Which simplifies to e^{-4} or \frac{1}{e^4} and you should leave it in this form.
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    I was just doing this last night but I guess your question has been answered now. I went a different way about it personally but the result was the same.
 
 
 
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