Checking MVT hypotheses for a piecewise function

First check continuity at $x=1$, the only point where the formula changes. From the left, $\lim_{x\to 1^-} x^2 = 1$; from the right, $\lim_{x\to 1^+} (2x-1) = 1$; and $f(1) = 1^2 = 1$. Since all three values agree, $f$ is continuous at $x=1$, and since both pieces are polynomials elsewhere, $f$ is continuous on all of $[0,2]$.

Next check differentiability at $x=1$. For $x \le 1$, we have $f'(x) = 2x$, so the left-hand derivative at $x=1$ is $f'_-(1) = 2(1) = 2$. For $x > 1$, we have $f'(x) = 2$, so the right-hand derivative at $x=1$ is $f'_+(1) = 2$. Since both one-sided derivatives equal $2$, $f$ is differentiable at $x=1$, so $f$ is differentiable on $(0,2)$.

Since $f$ is continuous on $[0,2]$ and differentiable on $(0,2)$, both hypotheses of the Mean Value Theorem are satisfied. Therefore MVT guarantees there exists some $c \in (0,2)$ such that $f'(c) = \dfrac{f(2)-f(0)}{2-0} = \dfrac{3-0}{2} = \dfrac{3}{2}$.

The answer is 2. At the junction point $x=1$, the left-hand derivative of $f$ is $f'_-(1) = 2$ (since $f'(x) = 2x$ for $x \le 1$), and the right-hand derivative of $f$ is $f'_+(1) = 2$ (since $f'(x) = 2$ for $x > 1$). Since these one-sided derivatives matched, $f$ was differentiable at that point, which was essential for confirming the Mean Value Theorem's hypotheses held on the whole interval.