Question 875141
critical numbers of y=3x^3-12x^2-7 
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begin by finding the y'
y=3x^3-12x^2-7
y'=9x^2-24x
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Set y' to zero and solve for x to find critical points:
0=9x^2-24x
0=(3x)(3x-8)
x = {0, 8/3}  (critical points)