find the points on the curve y= 2x^3+ 3x^2 - 12x + 1 where the tangent is horizontal

Question

find the points on the curve y= 2x^3 + 3x^2 - 12x + 1 where the tangent is horizontal 

Answer 1

Set the derivative of y equal to zero and solve for x.

 

y' = 0

 

6x² + 6x - 12 = 0

 

6(x² + x - 2) = 0

 

6(x + 2)(x - 1) = 0

 

x = -2          and            x = 1

 

Since a cubic function has 2 critical points, horizontal tangents occur at x=-2 and x=1.