Question

Let $f$ be differentiable on $(a, b)$ and continuous on $[a, b]$. Then there is at least one $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$. This is the statement of}

• A. Fundamental Theorem of Calculus
• B. Fundamental Theorem of Algebra
• C. Mean-Value Theorem
• D. Rolle's Theorem

Hint:

Rolle's Theorem is simply the Mean Value Theorem with the added constraint $f(a) = f(b)$, which makes the average rate of change $\frac{f(b)-f(a)}{b-a} = 0$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This theorem links the average rate of change of a continuous, differentiable function over an interval to the instantaneous rate of change (derivative) at some point within that interval.

Step 2: Detailed Explanation:
Let us review the mathematical statements of the options:
- Fundamental Theorem of Calculus: Connects differentiation and integration, stating that $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
- Fundamental Theorem of Algebra: States that every non-zero single-variable polynomial of degree $n$ has exactly $n$ complex roots.
- Rolle's Theorem: A special case of the Mean Value Theorem where $f(a) = f(b)$, concluding that there is some $c \in (a, b)$ where $f'(c) = 0$.
- Mean-Value Theorem (specifically Lagrange's Mean Value Theorem): States that if a function $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one number $c$ in the open interval $(a, b)$ such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Geometrically, this means there is at least one point where the secant line joining the endpoints is parallel to the tangent line at $c$.
Therefore, the statement is the Mean-Value Theorem.

Step 3: Final Answer
The correct option is (C).