Question:

Step 1: Express $f(x)$, $f'(x)$, and $f''(x)$ in terms of unknown constants. Let $A = g'(1)$ and $B = g''(2)$. The function $f(x)$ is given by $f(x) = x^2 + g'(1)x + g''(2)$. Substituting the constants, we have: $$f(x) = x^2 + Ax + B$$ Now, we find the first and second derivatives of $f(x)$: $$f'(x) = \frac{d}{dx}(x^2 + Ax + B) = 2x + A$$ $$f''(x) = \frac{d}{dx}(2x + A) = 2$$ Step 2: Substitute $f(1)$, $f'(x)$, and $f''(x)$ into the expression for $g(x)$. The function $g(x)$ is given by $g(x) = f(1)x^2 + xf'(x) + f''(x)$. First, calculate $f(1)$ from $f(x) = x^2 + Ax + B$: $$f(1) = (1)^2 + A(1) + B = 1 + A + B$$ Now substitute $f(1)$, $f'(x)$, and $f''(x)$ into the expression for $g(x)$: $$g(x) = (1 + A + B)x^2 + x(2x + A) + 2$$ $$g(x) = (1 + A + B)x^2 + 2x^2 + Ax + 2$$ Combine the $x^2$ terms: $$g(x) = (1 + A + B + 2)x^2 + Ax + 2$$ $$g(x) = (3 + A + B)x^2 + Ax + 2$$ Step 3: Calculate $g'(x)$ and $g''(x)$. Differentiate $g(x)$ with respect to $x$ to find $g'(x)$: $$g'(x) = \frac{d}{dx}((3 + A + B)x^2 + Ax + 2) = 2(3 + A + B)x + A$$ Differentiate $g'(x)$ with respect to $x$ to find $g''(x)$: $$g''(x) = \frac{d}{dx}(2(3 + A + B)x + A) = 2(3 + A + B)$$ Step 4: Use the definitions of $A$ and $B$ to form equations and solve for the constants. Recall that $A = g'(1)$ and $B = g''(2)$. Using $A = g'(1)$: Substitute $x=1$ into $g'(x) = 2(3 + A + B)x + A$: $$A = 2(3 + A + B)(1) + A$$ $$A = 6 + 2A + 2B + A$$ $$A = 6 + 3A + 2B$$ Subtract $A$ from both sides: $$0 = 6 + 2A + 2B \quad (*)$$ Using $B = g''(2)$: Substitute $x=2$ (though $g''(x)$ is constant) into $g''(x) = 2(3 + A + B)$: $$B = 2(3 + A + B)$$ $$B = 6 + 2A + 2B$$ Rearrange the terms: $$-6 = 2A + B \quad (**)$$ Now we have a system of two linear equations for $A$ and $B$: $(*) \quad 2A + 2B = -6$ $(**) \quad 2A + B = -6$ Subtract equation $(**)$ from equation $(*)$: $$(2A + 2B) - (2A + B) = -6 - (-6)$$ $$B = 0$$ Substitute $B=0$ into equation $(**)$: $$2A + 0 = -6$$ $$2A = -6$$ $$A = -3$$ So, the constants are $A = -3$ and $B = 0$. Step 5: Determine the explicit forms of $f(x)$ and $g(x)$. Using $A = -3$ and $B = 0$: $$f(x) = x^2 + Ax + B = x^2 - 3x + 0$$ $$f(x) = x^2 - 3x$$ Now, substitute $A=-3$ and $B=0$ into the expression for $g(x)$: $$g(x) = (3 + A + B)x^2 + Ax + 2$$ $$g(x) = (3 + (-3) + 0)x^2 + (-3)x + 2$$ $$g(x) = (0)x^2 - 3x + 2$$ $$g(x) = -3x + 2$$ Step 6: Calculate $f(4)$ and $g(4)$. Substitute $x=4$ into $f(x) = x^2 - 3x$: $$f(4) = (4)^2 - 3(4) = 16 - 12 = 4$$ Substitute $x=4$ into $g(x) = -3x + 2$: $$g(4) = -3(4) + 2 = -12 + 2 = -10$$ Step 7: Compute the value of $f(4) - g(4)$. $$f(4) - g(4) = 4 - (-10)$$ $$f(4) - g(4) = 4 + 10$$ $$f(4) - g(4) = 14$$ The final answer is $\boxed{14}$.