Rolle's theorem holds for monic quadratic polynomial $f(x)$ on the interval $[\alpha, \alpha + 3]$ where $f(\alpha) = 0$. Similarly, $g(x) = f(x) + 2$ also follows Rolle's theorem in the interval $[\beta, 3]$ where $g(3) = 0$, such that the value of $c$ is the same for both $f(x)$ and $g(x)$. Then the value of $(f \circ g)(\alpha)$ is...

Since $f(x)$ is a monic quadratic polynomial and Rolle's theorem holds on $[\alpha, \alpha + 3]$ with $f(\alpha) = 0$, we have $f(\alpha + 3) = f(\alpha) = 0$.

Thus, $f(x) = (x - \alpha)(x - \alpha - 3)$.

The value of $c$ for $f(x)$ is the midpoint of the interval, $c = \alpha + \dfrac{3}{2}$.

For $g(x) = f(x) + 2$, Rolle's theorem holds on $[\beta, 3]$. The value of $c$ is the same, so the midpoint of $[\beta, 3]$ is also $\alpha + \dfrac{3}{2}$.

$\dfrac{\beta + 3}{2} = \alpha + \dfrac{3}{2} \Rightarrow \beta = 2\alpha$.

We are given $g(3) = 0$, which means $f(3) + 2 = 0 \Rightarrow f(3) = -2$.

Substituting $x = 3$ in $f(x)$, we get:

$(3 - \alpha)(3 - \alpha - 3) = -2 \Rightarrow -\alpha(3 - \alpha) = -2 \Rightarrow \alpha^2 - 3\alpha + 2 = 0$.

This gives $\alpha = 1$ or $\alpha = 2$.

If $\alpha = 2$, then $\beta = 4$, making the interval $[4, 3]$, which is invalid. Thus, $\alpha = 1$ and $\beta = 2$.

So, $f(x) = (x - 1)(x - 4)$ and $g(x) = f(x) + 2$.

We need to find $(f \circ g)(\alpha) = f(g(1))$.

Since $f(1) = 0$, $g(1) = f(1) + 2 = 2$.

Then, $f(g(1)) = f(2) = (2 - 1)(2 - 4) = -2$.

Answer: $-2$