What is the extreme point of the curve 3x-6x^2?
We have to find the extreme point of the curve y = 3x - 6x^2.
To do that we find the first derivative of 3x - 6x^2 and equate it to zero. This is solved for x.
Now y = 3x - 6x^2
y' = 3 - 12x
3 - 12x = 0
=> 12x = 3
=> x = 1/4
At x = 1/4, y = 3*(1/4) - 6*(1/4)^2
=> 3/4 - 6/16
=> 3/4 -3/8
Also at x = 1/4, y'' is -12 which is negative. So we have the point of maxima at x = 1/4
The extreme point is at x = 1/4 and this is the maximum point with the expression equal to 3/8.
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