Animated Solution for Mathematics - Differentiation: If $f(x)=x^2+g^{\prime }(1)x+g^{\prime \prime }(2)$ and $g(x)=f(1)x^2+x{f}^{\prime }(x)+f^{\prime \prime }(x)$, then the value of $f(4)-g(4)$ is equal to ______.

Imagine you are standing before a mirror, and that mirror is reflecting another mirror. That is exactly what we have here with $f(x)$ and $g(x)$. They are intertwined, each defined by the other's behavior.

We are given:

The secret to solving this is to stop seeing $g^{\prime }(1)$ and $g^{\prime \prime }(2)$ as scary symbols and start seeing them as what they truly are: constants. Let us define $A=g^{\prime }(1)$ and $B=g^{\prime \prime }(2)$. Suddenly, the problem becomes a puzzle of finding two numbers.

The Power of Differentiation

We need to know what $f^{\prime }(x)$ and $f^{\prime \prime }(x)$ look like to understand $g(x)$. Given $f(x)=x^2+Ax+B$, the derivatives are:

This is the breakthrough! The second derivative of $f$ is just a constant, $2$. This simplifies our expression for $g(x)$ immensely.

Substituting these into the definition of $g(x)$:

The Unraveling

To find our constants, we differentiate $g(x)$:

Now, evaluate this at $x=1$. Since we defined $g^{\prime }(1)=A$, we have:

The $A$ on both sides cancels out perfectly, leaving $0=2(f(1)+2)$. This tells us that $f(1)=-2$.

Next, we find $g^{\prime \prime }(2)$. From our expression for $g^{\prime }(x)$, we know $g^{\prime \prime }(x)=2(f(1)+2)$. Substituting $f(1)=-2$: Thus, $g^{\prime \prime }(2)=0$.

The Final Construction

We have $f(1)=-2$ and $g^{\prime \prime }(2)=0$. Returning to our original definition of $f(x)$:

Now we have the complete functions:

Finally, we calculate the values at $x=4$:

The difference is:

You untangled the system, navigated the calculus, and arrived at the elegant solution of 14.