Find the points on the curve y=2x^3-9x^2+42 where the tangent like is horizontal?

Question

I understand how to take the derivative first and I get y'=6x^2-18x but I don't know what to do from there :(

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Answer 1 · 2018-03-06T16:00:59

#x=0# and #x=3#

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All you need do now is set y'=0, giving you the quadratic equation 6x²-18x+0=0, which has solutions #x=0# and #x=3#.