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I need this solved for x:
y' = 4ax^3 + 3bx^2 + 2cx + d = 0
This is to say, I need the formula for the "critical points" of a Quartic function.
Wikipedia says: "The derivative of a quartic function is a cubic function."
https://en.wikipedia.org/wiki/Quartic_function
And I found the above derivative here:
http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Glazer/essays/HTML/quartinfsola.html
Any help solving it would be great (it is beyond my ability).
Thanks
y' = 4ax^3 + 3bx^2 + 2cx + d = 0
This is to say, I need the formula for the "critical points" of a Quartic function.
Wikipedia says: "The derivative of a quartic function is a cubic function."
https://en.wikipedia.org/wiki/Quartic_function
And I found the above derivative here:
http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Glazer/essays/HTML/quartinfsola.html
Any help solving it would be great (it is beyond my ability).
Thanks