1. Find Solution using Lagrange's Inverse Interpolation formula

x	f(x)
168	3
120	7
72	9
63	10

x = 6
Finding f(2)

Solution:
The value of table for `x` and `y`

x	168	120	72	63
y	3	7	9	10

Lagrange's Inverse Interpolating Polynomial
The value of y at you want to find `P_n(y) : y = 6`

Lagrange's Inverse Interpolation formula is
`f(y) = ((y - y_1)(y - y_2)(y - y_3))/((y_0 - y_1)(y_0 - y_2)(y_0 - y_3)) xx x_0 + ((y - y_0)(y - y_2)(y - y_3))/((y_1 - y_0)(y_1 - y_2)(y_1 - y_3)) xx x_1 + ((y - y_0)(y - y_1)(y - y_3))/((y_2 - y_0)(y_2 - y_1)(y_2 - y_3)) xx x_2 + ((y - y_0)(y - y_1)(y - y_2))/((y_3 - y_0)(y_3 - y_1)(y_3 - y_2)) xx x_3`

`x(6) = ((6 - 7)(6 - 9)(6 - 10))/((3 - 7)(3 - 9)(3 - 10)) xx 168 + ((6 - 3)(6 - 9)(6 - 10))/((7 - 3)(7 - 9)(7 - 10)) xx 120 + ((6 - 3)(6 - 7)(6 - 10))/((9 - 3)(9 - 7)(9 - 10)) xx 72 + ((6 - 3)(6 - 7)(6 - 9))/((10 - 3)(10 - 7)(10 - 9)) xx 63`

`x(6) = ((-1)(-3)(-4))/((-4)(-6)(-7)) xx 168 + ((3)(-3)(-4))/((4)(-2)(-3)) xx 120 + ((3)(-1)(-4))/((6)(2)(-1)) xx 72 + ((3)(-1)(-3))/((7)(3)(1)) xx 63`

`x(6) = 0.0714 xx 168 + 1.5 xx 120 + (-1) xx 72 + 0.4286 xx 63`

`x(6) = 147`


Solution of the polynomial at point `6` is `x(6) = 147`