Repeat Exercise 21 for \({{\bf{f}}_0}\left( {\frac{3}{4}} \right)\), \({{\bf{f}}_1}\left( {\frac{3}{4}} \right)\), and \({\bf{g}}\left( {\frac{3}{4}} \right)\).

It is given that \({{\rm{p}}_0},{\rm{ }}{{\rm{p}}_1},{\rm{ }}{{\rm{p}}_2} \in {\mathbb{R}^n}\). Also,\({{\rm{f}}_0}\left( t \right) = \left( {1 - t} \right){{\rm{p}}_0} + t{{\rm{p}}_1}\), \({{\rm{f}}_1}\left( t \right) = \left( {1 - t} \right){{\rm{p}}_1} + t{{\rm{p}}_2}\)and \({\rm{g}}\left( t \right) = \left( {1 - t} \right){{\rm{f}}_0}\left( t \right) + t{{\rm{f}}_1}\left( t \right)\).The graph of\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\),\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\) and\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\) is to be drawn.

The values of \({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\),\({{\rm{f}}_1}\left( {\frac{3}{4}} \right)\) and\({\rm{g}}\left( {\frac{3}{4}} \right)\)are calculated as:

\(\begin{aligned}{}{{\rm{f}}_0}\left( {\frac{3}{4}} \right) &= \left( {1 - \frac{3}{4}} \right){{\rm{p}}_0} + \frac{3}{4}{{\rm{p}}_1}\\ &= \frac{1}{4}{{\rm{p}}_0} + \frac{3}{4}{{\rm{p}}_1}\\\frac{1}{4}\left( {{{\rm{p}}_0} + 3{{\rm{p}}_1}} \right)\end{aligned}\)

It shows that\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\)is\(\frac{1}{4}\)of the distance of the line\({{\rm{p}}_0}{\rm{, }}{{\rm{p}}_1}\)from the point\({{\rm{p}}_0}\), and\(\frac{3}{4}\)of the distance of the line\({{\rm{p}}_0}{\rm{, }}{{\rm{p}}_1}\)from the point\({{\rm{p}}_1}\).

\(\begin{aligned}{}{{\rm{f}}_1}\left( {\frac{3}{4}} \right) &= \left( {1 - \frac{3}{4}} \right){{\rm{p}}_1} + \frac{3}{4}{{\rm{p}}_2}\\ &= \frac{1}{4}{{\rm{p}}_1} + \frac{3}{4}{{\rm{p}}_2}\\\frac{1}{4}\left( {{{\rm{p}}_1} + 3{{\rm{p}}_2}} \right)\end{aligned}\)

It shows that \({{\rm{f}}_1}\left( {\frac{3}{4}} \right)\)is\(\frac{1}{4}\)of the distance of the line\({{\rm{p}}_0}{\rm{ }}{{\rm{p}}_1}\)from the point\({{\rm{p}}_0}\), and\(\frac{3}{4}\)of the distance of the line\({{\rm{p}}_0}{\rm{ }}{{\rm{p}}_1}\)from the point\({{\rm{p}}_1}\).

\(\begin{aligned}{}{\rm{g}}\left( {\frac{3}{4}} \right) &= \left( {1 - \frac{3}{4}} \right){{\rm{f}}_0}\left( {\frac{3}{4}} \right) + \frac{1}{2}{{\rm{f}}_1}\left( {\frac{3}{4}} \right)\\ &= \frac{1}{4}{{\rm{f}}_0}\left( {\frac{3}{4}} \right) + \frac{3}{4}{{\rm{f}}_1}\left( {\frac{3}{4}} \right)\\ &= \frac{1}{4}\left( {{{\rm{f}}_0}\left( {\frac{3}{4}} \right) + 3{{\rm{f}}_1}\left( {\frac{3}{4}} \right)} \right)\end{aligned}\)

It shows that \({\rm{g}}\left( {\frac{3}{4}} \right)\)is\(\frac{1}{4}\)of the distance of the line\({{\rm{f}}_0}\left( {\frac{3}{4}} \right){{\rm{f}}_1}\left( {\frac{3}{4}} \right)\)from the point\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\), and\(\frac{3}{4}\)of the distance of the line\({{\rm{p}}_0}{\rm{ }}{{\rm{p}}_1}\)from the point\({{\rm{f}}_1}\left( {\frac{3}{4}} \right)\).

Draw lines joining points\({{\rm{p}}_0}{\rm{, }}{{\rm{p}}_1}\), and points\({{\rm{p}}_1}{\rm{, }}{{\rm{p}}_2}\),such that\({{\rm{f}}_0}\left( {\frac{3}{4}} \right)\),\({{\rm{f}}_1}\left( {\frac{3}{4}} \right)\)is one fourth from front end and three fourth from rear end.

Afterwards, the\({\rm{g}}\left( {\frac{3}{4}} \right)\)is at its one fourth from front end and three fourth from rear end of the line joining\({{\rm{f}}_0}\left( {\frac{1}{2}} \right)\),\({{\rm{f}}_1}\left( {\frac{1}{2}} \right)\).

The diagram is shown below: