Question

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 ______

Answer 1

Correct Answer: 14

The correct answer is 14.
$f(x)=x^2+g^{\prime }(1)x+g^{\prime \prime }(2)$
$f^{\prime }(x)=2x+g^{\prime }(1)$
$f^{\prime \prime }(x)=2$
$g(x)=f(1)x^2+x\left[2x+g^{\prime }(1)\right]+2$
$g^{\prime }(x)=2f(1)x+4x+g^{\prime }(1)$
$g^{\prime \prime }(x)=2f(1)+4$
$g^{\prime \prime }(x)=0$
$2f(1)+4=0$
$f(1)=-2$
$-2=1+g^{\prime }(1)=g^{\prime }(1)=-3$
So $f^{\prime }(x)=2x-3$
$f(x)=x^2-3x+c$
$c=0$
$f(x)=x^2-3x$
$g(x)=-3x+2$
$f(4)-g(4)=14$

Answer 2

Computing derivatives: \[ f'(x) = 2x + g'(1), \quad f''(x) = 2 \] Solving for \( g(x) \), \[ g(x) = f(1)x^2 + x[2x + g'(1)] + 2 \] \[ g'(x) = 2f(1)x + 4x + g'(1) \] \[ g''(x) = 2f(1) + 4 \] Setting boundary conditions: \[ f(1) = -2, \quad g'(1) = -3 \] \[ f(x) = x^2 - 3x \] \[ g(x) = -3x + 2 \] \[ f(4) - g(4) = 14 \]