What is the smallest possible slope for y=x^3 -3x^2 +5x -1

let's find its first derivative for "y"

now, we can run the quadratic formula on that derivative, and it turns out we end up with a negative radicand and therefore an imaginary/complex solution, which is a way to say there are no solutions for it, so the quadratic never touches the x-axis.

so hmmm what is the smallest value it goes down to, for the derivative, before it goes right back up?

well, for that we can simply check her vertex location.

so, "y" or dy/dx goes as low as 2 before going back up, and that happens when x = 1.

Check the picture below.

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step-by-step explanation: