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**Lagrange's Mean Value Theorem (MVT)** states: for $f$ continuous on $[a,b]$, differentiable on $(a,b)$, there exists $c \in (a, b)$ such that:
A$f'(c) = 0$
B$f'(c) = \dfrac{f(b) - f(a)}{b - a}$
C$f(c) = 0$
D$f''(c) = 0$
Answer & Solution
Correct answer: B. $f'(c) = \dfrac{f(b) - f(a)}{b - a}$
MVT: there exists $c$ where instantaneous rate of change $f'(c)$ equals the average rate $\dfrac{f(b)-f(a)}{b-a}$. Rolle's theorem is the special case where $f(a) = f(b)$.
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