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    Name:  IMG_3163.jpg
Views: 48
Size:  337.7 KBShow that, for all non-zero values of the constant m, the curve
    y=e^mx (x^2 +mx)
    has exactly two stationary points.


    so far, I've found that dy/dx => e^mx (mx^2 +xm^2 +2x+m)=0
    but I'm not sure where to go from here
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    Use the discriminant.
    For a curve to have 2 stationary points, dy/dx=0 will have distinct real roots.
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    (Original post by B_9710)
    Use the discriminant.
    For a curve to have 2 stationary points, dy/dx=0 will have distinct real roots.
    Name:  IMG_3164.jpg
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    I've done this, but how do I use this to prove the stationary points?
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    Please tell me this isn't C3 or c4?
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    (Original post by zayn008)
    Please tell me this isn't C3 or c4?
    This is C3
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    (Original post by Lollieboo)
    This is C3
    OMG This year is gonna be horrible ☹️
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    (Original post by Lollieboo)
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    I've done this, but how do I use this to prove the stationary points?
    m^4 is always positive so youve shown b^2-4ac>0 so youre done
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    (Original post by newblood)
    m^4 is always positive so youve shown b^2-4ac>0 so youre done
    phew, THANK YOU
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    well m^4 will always be positive (it's m^2 x m^2)
    + 4 is always gonna be positive as well, so I think you're there pal
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    (Original post by Lollieboo)
    Name:  IMG_3164.jpg
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    I've done this, but how do I use this to prove the stationary points?
    You've already proven it. The fact your discriminant is equal to m^4 + 4 is enough proof.*

    No matter what value of M you input, that equation will always produce a positive value which is above 0. *
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    (Original post by zayn008)
    OMG This year is gonna be horrible ☹️
    Its not exactly that difficult at all. Its just product rule of differentiation, practice it enough and this stuff will become second nature. The only difficulty here is remembering the formula.
 
 
 
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