If $$f(x) = x^2 + g'(1)x + g''(2)$$ and $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$, then the value of $$f(4) - g(4)$$ is equal to _____.

Given: $$f(x) = x^2 + g'(1)x + g''(2)$$ and $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$.

Let $$a = g'(1)$$ and $$b = g''(2)$$. Then $$f(x) = x^2 + ax + b$$.

Computing f values:

$$f(1) = 1 + a + b$$, $$f'(x) = 2x + a$$, $$f''(x) = 2$$

Substituting into g:

Computing g values:

$$g'(x) = 2(3+a+b)x + a$$, so $$g'(1) = 2(3+a+b) + a = 6 + 3a + 2b$$

$$g''(x) = 2(3+a+b)$$, so $$g''(2) = 2(3+a+b)$$

Solving the equations:

From $$g'(1) = a$$: $$6 + 3a + 2b = a \Rightarrow 2a + 2b = -6 \Rightarrow a + b = -3$$ ... (1)

From $$g''(2) = b$$: $$b = 2(3 + a + b) = 2(3 - 3) = 0$$ (using (1))

So $$b = 0$$ and $$a = -3$$.

$$f(x) = x^2 - 3x$$, $$g(x) = (3-3+0)x^2 + (-3)x + 2 = -3x + 2$$

Final computation:

Therefore, $$f(4) - g(4) = 14$$.