To find the critical points we have to take the derivative f'(x)=6x^2-72x+120 and set it equal to 0 i.e.
x^2-12x+20=0 which has two zeros at x=2 and x=10.
Joe S.
asked • 10/28/20calc dont know how to do it.
The function f(x)=2x^3−36x^2+120x−10 has two critical numbers.
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