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d/dt(3t^2- 2t^(-3))= d/dt(3t^2)- d/dt(2t^(-3))= 6t- (-6t^(-4))= 6t+ 6t^(-4).
The horizontal tangents are the tangents whose slope is zero. To find the tangent we have:
dy/dx= 3x^2+ 1
The question is: Can dy/dx= 0 happen here?
3x^2+ 1=0 ---> x^2= -1/3---> no roots. No ...
This shows how to find a derivative, the number of horizontal tangents for a given polynomial, and marginal cost.