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At what points on the curve f(x)=2-4x^3 does the normal have a slope of 3?

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if the slope of the normal is 3 , then the slope of the tangent is -1/3

f'(x) = -12 x^2

-12 x^2 = -1/3 ... x^2 = 1/36 ... x = ± 1/6

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