I'm given a partial table from which to work:
$x_0=0$ $P_0=1$
$x_1=0.25$ $P_1=2$ $P_{0,1}=2.6$
$x_2=0.5$ $P_2= $ $P_{1,2}= $ $P_{0,1,2}=$
$x_3=0.75$ $P_3=8$ $P_{2,3}=2.4$ $P_{1,2,3}=2.96$ $P_{1,2,3,4}=3.016$
I'm thinking I can find it recursively by
$P_{2,3}= \dfrac{(x-x_2)P_3-(x-x_3)P_2}{x_3-x_2}$ and solving for $P_2$
I got:
$2.4=\dfrac{(0.4-0.5)*8-(0.4-0.75)P_2}{(.75-.5)}$
And got $P_2=4$. The answer given in the back of the book is $4$.
So I think I did it right.
Thanks!