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Degree 4 Gauss–Lobatto–Legendre on a interval

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

\(\displaystyle \phi_{0} = \frac{32 x^{4}}{3} - \frac{80 x^{3}}{3} + \frac{70 x^{2}}{3} - \frac{25 x}{3} + 1\)

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

\(\displaystyle \phi_{1} = \frac{x \left(32 x^{3} - 48 x^{2} + 22 x - 3\right)}{3}\)

This DOF is associated with vertex 1 of the reference element.
\(\displaystyle l_{2}:v\mapsto v(\tfrac{1}{4})\)

\(\displaystyle \phi_{2} = \frac{16 x \left(- 8 x^{3} + 18 x^{2} - 13 x + 3\right)}{3}\)

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

\(\displaystyle \phi_{3} = 4 x \left(16 x^{3} - 32 x^{2} + 19 x - 3\right)\)

This DOF is associated with edge 0 of the reference element.
\(\displaystyle l_{4}:v\mapsto v(\tfrac{3}{4})\)

\(\displaystyle \phi_{4} = \frac{16 x \left(- 8 x^{3} + 14 x^{2} - 7 x + 1\right)}{3}\)

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