let's find its first derivative for "y"
[tex]\bf y=x^3-3x^2+5x-1\implies \cfrac{dy}{dx}=3x^2-6x+5[/tex]
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.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \cfrac{dy}{dx}=\stackrel{\stackrel{a}{\downarrow }}{3}x^2\stackrel{\stackrel{b}{\downarrow }}{-6}x\stackrel{\stackrel{c}{\downarrow }}{+5} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{(-6)}{2(3)}~~,~~5-\cfrac{(-6)^2}{4(3)} \right)\implies \left( 1~~,5-\cfrac{36}{12} \right) \\\\\\ (1~,~5-3)\implies (\stackrel{x}{1}~,~\stackrel{y}{2})[/tex]
so, "y" or dy/dx goes as low as 2 before going back up, and that happens when x = 1.
Check the picture below.
