MHB Mean Value Theorem: Showing Change in a Function is Bounded

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The Mean Value Theorem (MVT) asserts that for a function f continuous on [2,8] and differentiable on (2,8), there exists at least one point x in (2,8) where the instantaneous rate of change f'(x) equals the average rate of change over the interval. The average rate of change is calculated as (f(8) - f(2)) / (8 - 2). Given that the derivative f'(x) is bounded between 3 and 5, it follows that the change in the function values, f(8) - f(2), must lie between 18 and 30. This demonstrates how the MVT provides bounds on the change of a function even without a specific function defined. The discussion highlights the application of the theorem to derive meaningful conclusions about the behavior of functions over specified intervals.
karush
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Ok Just have trouble getting this without a function..
 
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average rate of change of $f(x)$ on the interval $[2,8]$ is $\dfrac{f(8)-f(2)}{8-2}$

the MVT states there exists at least one value of $x \in (2,8)$ where $f'(x) = \dfrac{f(8)-f(2)}{8-2}$

$3 \le f'(x) \le 5 \implies 3 \le \dfrac{f(8)-f(2)}{8-2}\le 5 \implies 18 \le f(8)-f(2) \le 30$
 

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