Question 998301
Determine the location of each extremum of the function
f(x)= x^3-1/2x^2-6x+4
f(x)= 3x^2-x-6 ( = f'(x), not f(x) )
    = x^2-x/3-2=0 that's all i got
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Set f' = 0
3x^2 - x - 6 = 0
The extrema are at the zeroes of f'(x)
*[invoke solve_quadratic_equation 3,-1,-6]
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Those are the x values for the max (negative value) and min (positive value).
Finding the y values is tedious, but just algebra.