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Difficult Differentiation Question

IMG_3163.jpgShow 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
(edited 7 years ago)
Reply 1
Use the discriminant.
For a curve to have 2 stationary points, dy/dx=0 will have distinct real roots.
(edited 7 years ago)
Reply 2
Original post by B_9710
Use the discriminant.
For a curve to have 2 stationary points, dy/dx=0 will have distinct real roots.


IMG_3164.jpg
I've done this, but how do I use this to prove the stationary points?
Reply 3
Please tell me this isn't C3 or c4?
Reply 4
Original post by zayn008
Please tell me this isn't C3 or c4?


This is C3 :frown:
Reply 5
Original post by Lollieboo
This is C3 :frown:


OMG This year is gonna be horrible ☹️
Original post by Lollieboo
IMG_3164.jpg
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
Reply 7
Original post by newblood
m^4 is always positive so youve shown b^2-4ac>0 so youre done


phew, THANK YOU
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
Reply 9
IMG_3165.jpg
Original post by Lollieboo
IMG_3164.jpg
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. *
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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