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    bonjour!
    please help if you have the know how

    y= x.e^2x
    find the turning point an determine it's nature.


    I know I need to find dy/dx, set it equal to 0, solve for x and then plug that in to find y. I think I know how to determine nature (second diff?)

    I'm having trouble with the numbers
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    Ok, what have you done so far? Have you found \frac{dy}{dx}?
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    (Original post by Noble.)
    Ok, what have you done so far? Have you found \frac{dy}{dx}?
    I think it's 2x.e^2x but I had trouble solving for x when that =0
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    (Original post by pete.mcfc)
    bonjour!
    please help if you have the know how

    y= x.e^2x
    find the turning point an determine it's nature.


    I know I need to find dy/dx, set it equal to 0, solve for x and then plug that in to find y. I think I know how to determine nature (second diff?)

    I'm having trouble with the numbers
    you should have found the derivative: 2x.e^(2x) + e^(2x)

    Set this equal to zero, factorise out the e^(2x) term ... and hopefully you will see that x = -0.5


    is this what you have so far??
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    (Original post by pete.mcfc)
    I think it's 2x.e^2x but I had trouble solving for x when that =0
    Unfortunately you have forgotten to use the product rule.. Give it another shot !!
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    (Original post by Mrm.)
    Unfortunately you have forgotten to use the product rule.. Give it another shot !!
    Thanks, I can do the product rule but I always try and differentiate without it!
    Nice one got it now
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    (Original post by pete.mcfc)
    I think it's 2x.e^2x but I had trouble solving for x when that =0
    Remember the product rule.

    \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
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    (Original post by Noble.)
    Remember the product rule.

    \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
    nice one, got that bit.

    if dy/dx is 2x.e^2x + e^2x and x= 0.5 and y= 0.184 do i find d2y/dx2 and plug in x to determine nature?
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    (Original post by pete.mcfc)
    nice one, got that bit.

    if dy/dx is 2x.e^2x + e^2x and x= 0.5 and y= 0.184 do i find d2y/dx2 and plug in x to determine nature?
    \frac{dy}{dx} = 2xe^{2x} + e^{2x}

    Are you sure x=0.5 is a solution to \frac{dy}{dx} = 0?

    Factorise out the e^{2x} and knowing that e^{2x} \neq 0 you should find just one solution (you've got the magnitude correct, just not the sign )

    To find \frac{d^2y}{dx^2} you need to just apply the product rule again to 2xe^{2x}, then the rest is just straight differentiation.
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    This might help if you're still stuck.

    \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

    \frac{d}{dx}[f'(x)g(x) + f(x)g'(x)] = f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x)
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    (Original post by Noble.)
    \frac{dy}{dx} = 2xe^{2x} + e^{2x}

    Are you sure x=0.5 is a solution to \frac{dy}{dx} = 0?

    Factorise out the e^{2x} and knowing that e^{2x} \neq 0 you should find just one solution (you've got the magnitude correct, just not the sign )

    To find \frac{d^2y}{dx^2} you need to just apply the product rule again to 2xe^{2x}, then the rest is just straight differentiation.
    sorry I meant -0.5
    i tried to find the second diff without the product rule again!! i am a fool lol, but thaks a million for helping
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    (Original post by pete.mcfc)
    sorry I meant -0.5
    i tried to find the second diff without the product rule again!! i am a fool lol, but thaks a million for helping
    No problem.
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    (Original post by pete.mcfc)
    Thanks, I can do the product rule but I always try and differentiate without it!
    Nice one got it now
    no problem.
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    1) find dy /dx
    2) make dy /dx = 0, and solve for x values
    3) find d2y/ dx2, substitute x values in, and determine nature of SP.
 
 
 
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