What is the extreme point of the curve 3x-6x^2?
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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
=> 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.
Related Questions
To determine the extreme point of the curve, we'll have to calculate the critical point of the function, that is the root of the first derivative of f(x).
f'(x) = (3x-6x^2)'
f'(x) = 3 - 12x
We'll put f'(x) = 0.
3 - 12x = 0
We'll divide by 3:
1 - 4x = 0
We'll subtract 1:
-4x = -1
x = 1/4
The extreme value of the function is:
f(1/4) = 3/4 - 6/16
f(1/4) = (12-6)/16
f(1/4) = 6/16
We'll simplify and we'll get;
f(1/4) = 3/8
The extreme point of the curve is the vertex of the parabola and it is the maximum point of the function: (1/4 ; 3/8).
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