1. The problem asks to find the value of $c$ in Lagrange's Mean Value Theorem (MVT) for the function $f(x) = x(x - 1)$ on the interval $[1, 2]$. 2. Recall that Lagrange's MVT states there exists some $c \in (a, b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ where $a=1$ and $b=2$. 3. First, compute $f(1)$ and $f(2)$: $$f(1) = 1 \times (1 - 1) = 0$$ $$f(2) = 2 \times (2 - 1) = 2$$ 4. Calculate the average rate of change: $$\frac{f(2) - f(1)}{2 - 1} = \frac{2 - 0}{1} = 2$$ 5. Find the derivative $f'(x)$: $$f(x) = x^2 - x$$ $$f'(x) = 2x - 1$$ 6. Set $f'(c)$ equal to the average rate of change and solve for $c$: $$2c - 1 = 2$$ $$2c = 3$$ $$c = \frac{3}{2}$$ 7. The value of $c$ is $\frac{3}{2}$, which corresponds to option (b).