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h'(x) = 3x^2 - 6x + 2
Set $ h'(x) = 0 $ to find where the slope is horizontal:
3x^2 - 6x + 2 = 0
Solve using the quadratic formula:
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}
The sum of the roots is:
\left(1 + \frac{\sqrt{3}}{3}\right) + \left(1 - \frac{\sqrt{3}}{3}\right) = 2
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