Let f(x) and g(x) be twice differentiable functions on R satisfying...

Let $h(x) = f(x) - g(x)$.

We are given that $h''(x) = 24x$.

Integrating with respect to $x$, we get $h'(x) = 12x^2 + C$.

Since $h(x)$ has a local minimum at $x = 1$, we must have $h'(1) = 0$.

Substituting $x = 1$, $12(1)^2 + C = 0 \Rightarrow C = -12$.

Thus, $h'(x) = 12x^2 - 12$.

Integrating again with respect to $x$, we obtain $h(x) = 4x^3 - 12x + D$.

The minimum value at $x = 1$ is given as $-3$, so $h(1) = -3$.

Substituting $x = 1$, $4(1)^3 - 12(1) + D = -3 \Rightarrow -8 + D = -3 \Rightarrow D = 5$.

So, $h(x) = 4x^3 - 12x + 5$.

To find the local maximum, we set $h'(x) = 0$, which gives $12(x^2 - 1) = 0 \Rightarrow x = 1$ or $x = -1$.

Using the second derivative test, $h''(x) = 24x$. At $x = -1$, $h''(-1) = -24 < 0$, which confirms a local maximum at $x = -1$.

The local maximum value is $h(-1) = 4(-1)^3 - 12(-1) + 5 = -4 + 12 + 5 = 13$.

Answer: $13$