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Problem 121

# (a) State and prove the Mean Value Theorem for the derivative of a real valued function of a single real variable. (b) Give a geometrical interpretation to this result.

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The Mean Value Theorem states that if a function $$f(x)$$ is continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$, then there exists at least one point $$c \in (a, b)$$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}$$. In the proof, we define a new function $$g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)$$, show that it satisfies the conditions of Rolle's theorem, and use it to demonstrate the theorem's claim. The geometrical interpretation is that there exists a point within the given interval where the tangent line to the curve is parallel to the secant line connecting the points $$(a, f(a))$$ and $$(b, f(b))$$, meaning it has the same slope as the average rate of change between the interval's endpoints.
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## Step 1: Statement of Mean Value Theorem

The Mean Value Theorem states that: if a function $$f(x)$$ is continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$, then there exists at least one point $$c \in (a, b)$$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$

## Step 2: Proof of Mean Value Theorem

1. Let's define a new function $$g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)$$ 2. Then, $$g(a) = f(a) - \frac{f(b) - f(a)}{b - a}(0) = f(a)$$ and $$g(b) = f(b) - \frac{f(b) - f(a)}{b - a}(b - a) = f(a)$$ 3. Since $$f(x)$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b)$$, then $$g(x)$$ is also continuous on $$[a, b]$$ and differentiable on $$(a, b)$$ as it is a combination of continuous and differentiable functions. 4. Now, we apply Rolle's Theorem on $$g(x)$$, which states that if a function is continuous on $$[a, b]$$, differentiable on $$(a, b)$$, and $$g(a) = g(b)$$, then there exists at least one point $$c \in (a, b)$$ such that $$g'(c) = 0$$. 5. Differentiate $$g(x)$$ with respect to $$x$$: $$g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$$ 6. Since Rolle's Theorem states that $$g'(c) = 0$$, we plug in $$c$$ and solve for $$f'(c)$$: $$0 = f'(c) - \frac{f(b) - f(a)}{b - a}$$ 7. Therefore, $$f'(c) = \frac{f(b) - f(a)}{b - a}$$, proving the Mean Value Theorem. (b) Geometrical interpretation.

## Step 3: Interpretation

The Mean Value Theorem states that, given a continuous and differentiable function over an interval, there exists a point within that interval where the derivative (the tangent line) is equal to the average rate of change between the endpoints of the interval. In other words, there is a point in the interval where the tangent line to the curve is parallel to the secant line connecting the points $$(a, f(a))$$ and $$(b, f(b))$$. This theorem is extremely useful, as it allows us to understand how the function's derivative behaves over an interval and it also has numerous applications in calculus and real analysis.

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