Cubic polynomial interpolation

I'm going to use 2 methods to solve this problem, I'm going to give you a guideline for a third method (3), and of course, there are more methods ... for example, method of finite differences

$\boxed{1.-}$ We are looking for a cubic polynomial $p(x)$ fulfilling $p(1) = 1; \space p(2) = 2 ; \space p(3) = 3; \space p(4) = 5$ . I'm going to use Lagrange interpolation , and I'm going to search constants $a, b, c, d$ such that they meet the conditions imposed. $p(x) = a(x - 1)(x - 2)(x - 3) + b(x -1)(x -2)(x - 4) + c(x -2)(x - 3)(x - 4) + d(x - 1)(x - 3)(x - 4)$ $p(1) = 1 \Rightarrow c = \frac{-1}{6}; \space p(2) = 2 \Rightarrow d = 1; \space p(3) = 3 \Rightarrow b = \frac{-3}{2}; p(4) = 5 \Rightarrow a = \frac{5}{6}$ So, the polynomial is $p(x) = \frac{5}{6}(x -1)(x - 2)(x - 3) - \frac{3}{2}(x - 1)(x - 2)(x -4) - \frac{1}{6}(x - 2)(x - 3)(x - 4) + (x - 1)(x - 3)(x - 4)$ and now substituing at $x = 6$ we get $p(6) = 16$

$\boxed{2.-}$ If $p(x)$ is a cubic polynomial fulfilling the conditions imposed, then $p(x) - x$ is a cubic polynomial wich roots are $1,2,3$ and this implies that $p(x) - x = k(x - 1)(x- 2)(x- 3)$ where $k$ is a constant. $p(4) = 5 \Rightarrow k = \frac{1}{6} \Rightarrow p(x) = x + \frac{1}{6}(x - 1)(x - 2)(x - 3) \Rightarrow$ $p(6) = 6 + \frac{1}{6} \cdot 5 \cdot 4 \cdot 3 = 16$

$\boxed{3.-}$ Use method of indetermined coeffcients,i.e, look for a polynomial $p(x) = ax^3 + bx^2 + cx + d$ where $a, b ,c , d$ are constants and such that $p(1) = 1, \space p(2) = 2, \space, p(3) = 3, \space p(4) = 5$ ,i. e, look for constants $a ,b, c, d$ such that $a + b + c + d = 1 \\ 8a + 4b + 2c + d = 2 \\ 27a + 9b + 3c + d = 3 \\ 64a + 16b + 4c + d = 5$ Solve this system of equations for $a, b, c , d$ and later substitute...

$p(x) - x = \lambda(x-1)(x-2)(x-3)$ , where $\lambda$ is a leading coefficient .... $(1)$ .

Plug $x = 4$ to get

$p(4) = 6(\lambda) + 4$

$\large\ \lambda = \frac{1}{6}$ .

Put this is $(1)$ and set $x=6$ to get

$p(6) = 10 + 6 =16.$ .