1. Find Solution using Hermite's formula

x	f(x)
0	0
1	1

x = 0.1
Finding f(2)

Solution:
The value of table for `x`, `f(x)` and `f'(x)`

x	y=f(x)	y'=f'(x)
0	0	0
1	1	1

The Polynomials `I_i(x)` are

`I_0(x)=((x - x_1))/((x_0 - x_1))=((x -1))/((0 -1))=((x -1))/((-1))=(x-1)/(-1)=-x+1`

`I_1(x)=((x - x_0))/((x_1 - x_0))=((x -0))/((1 -0))=((x -0))/((1))=(x)/(1)=x`

`I_0'(x)=-1`

`I_1'(x)=1`

`I_0'(x_0)=I_0'(0)=-1=-1`

`I_1'(x_1)=I_1'(1)=1=1`

Hermite Interpolation Formula is
`H(x)=sum u_i(x)*y_i + sum v_i(x)*y_i'`

where `u_i(x)=[1-2(x-x_i) I_i'(x_i)][I_i(x)]^2` and `v_i(x)=(x-x_i)[I_i(x)]^2`

`u_0(x)=[1-2(x-x_0) I_0'(x_0)][I_0(x)]^2`

`=>u_0(x)=[1-2(x-0) I_0'(0)][I_0(x)]^2`

`=>u_0(x)=[1-2(x-0) * (-1)][I_0(x)]^2`

`=>u_0(x)=[1 +2x][I_0(x)]^2`

`=>u_0(x)=(2x+1)(-x+1)^2`

`v_0(x)=(x-x_0)[I_i(x)]^2`

`=>v_0(x)=(x)(-x+1)^2`

`u_1(x)=[1-2(x-x_1) I_1'(x_1)][I_1(x)]^2`

`=>u_1(x)=[1-2(x-1) I_1'(1)][I_1(x)]^2`

`=>u_1(x)=[1-2(x-1) * (1)][I_1(x)]^2`

`=>u_1(x)=[1 -2x+2][I_1(x)]^2`

`=>u_1(x)=(-2x+3)(x)^2`

`v_1(x)=(x-x_1)[I_i(x)]^2`

`=>v_1(x)=(x-1)(x)^2`

Hermite Interpolation formula is
`H(x)=u_0(x)*y_0+v_0(x)*y_0'+u_1(x)*y_1+v_1(x)*y_1'`

`H(x)=(2x+1)(-x+1)^2 * (0) + (x)(-x+1)^2 * (0)+(-2x+3)(x)^2 * (1) + (x-1)(x)^2 * (1)`

Putting x=0.1 and simplifying, we obtain
`H(0.1)=0.019`