An independent check uses Newton's divided-difference interpolation instead of Lagrange's form.
The zeroth divided differences are $f[x_0]=1$, $f[x_1]=2$, $f[x_2]=4$.
First divided differences:\[f[x_0,x_1]=\frac{f[x_1]-f[x_0]}{x_1-x_0}=\frac{2-1}{1-0}=1,\qquad f[x_1,x_2]=\frac{f[x_2]-f[x_1]}{x_2-x_1}=\frac{4-2}{3-1}=1\]
Second divided difference:\[f[x_0,x_1,x_2]=\frac{f[x_1,x_2]-f[x_0,x_1]}{x_2-x_0}=\frac{1-1}{3-0}=0\]
Newton's divided-difference polynomial is\[f(x)\approx f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)=1+1\cdot x+0=1+x\]
The vanishing second divided difference shows the three points actually lie on the straight line $f(x)=x+1$ (check: $f(0)=1$, $f(1)=2$, $f(3)=4$, all satisfied). So $f(2)\approx 1+2=3$, matching the Lagrange result exactly.
\[\boxed{f(2)\approx 3.00}\]