Guessing won't help

Let $f(x) = ax^3 + bx^2 + cx + d$

From the four pieces of information we form the equations:

$a+b+c+d = 1$

$-a+b-c+d = 11$

$-8a+4b-2c+d = 22$

$8a+4b+2c+d = 14$

Adding the first to the second and simplifying, we have $b+d = 6$ . Doing the same for the other two equations we have $4b+d = 18$ . Note that the question is looking for $f(0)$ which is simply the value of $d$ so we need only solve for $d$ .

$(4b+d) - (b+d) = 18-6$

$3b = 12$

$b=4$

$4+d = 6$

$d = 6-4 = \boxed{2}$ .