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Degree 3 Lagrange on a interval

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In this example:
\(\displaystyle l_{0}:v\mapsto v(0)\)

\(\displaystyle \phi_{0} = - 5 x^{3} + 10 x^{2} - 6 x + 1\)

This DOF is associated with vertex 0 of the reference element.
\(\displaystyle l_{1}:v\mapsto v(1)\)

\(\displaystyle \phi_{1} = x \left(5 x^{2} - 5 x + 1\right)\)

This DOF is associated with vertex 1 of the reference element.
\(\displaystyle l_{2}:v\mapsto v(\tfrac{1}{2} - \tfrac{\sqrt{5}}{10})\)

\(\displaystyle \phi_{2} = \frac{5 x \left(2 \sqrt{5} x^{2} - 3 \sqrt{5} x - x + 1 + \sqrt{5}\right)}{2}\)

This DOF is associated with edge 0 of the reference element.
\(\displaystyle l_{3}:v\mapsto v(\tfrac{\sqrt{5}}{10} + \tfrac{1}{2})\)

\(\displaystyle \phi_{3} = \frac{5 x \left(- 2 \sqrt{5} x^{2} - x + 3 \sqrt{5} x - \sqrt{5} + 1\right)}{2}\)

This DOF is associated with edge 0 of the reference element.