Checking MVT hypotheses on a discontinuous function

The Mean Value Theorem requires $f$ to be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Here $f(x) = \dfrac{1}{x-1}$ is undefined at $x=1$, since that makes the denominator zero, and $1$ lies inside $[0,3]$.

Because $f$ is not even defined (hence not continuous, and not differentiable) at $x=1 \in (0,3)$, the continuity hypothesis of the MVT fails on $[0,3]$. Therefore the theorem simply does not apply here, and no conclusion about the existence of such a $c$ can be drawn from it — the computation using $f(0)$ and $f(3)$ is meaningless in this context.

Indeed, $f$ has a vertical asymptote at $x=1$, splitting the graph into two disconnected branches on $[0,3]$, which visually confirms the discontinuity.