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What does the mean value theorem state for a function continuous on [a,b] and differentiable on (a,b)?

The mean value theorem states that if a function ff is continuous on the closed interval [a,b][a,b] and differentiable on the open interval (a,b)(a,b), then there exists at least one point cc in (a,b)(a,b) such that the instantaneous rate of change (derivative) at cc equals the average rate of change over the interval. Mathematically, this is expressed as f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b)-f(a)}{b-a}.

Conditions

  • a<ba < b
  • ff is continuous on [a,b][a,b]
  • ff is differentiable on (a,b)(a,b)

Reasoning, step by step

  1. Verify that the function is continuous on the closed interval [a,b][a,b].
  2. Verify that the function is differentiable on the open interval (a,b)(a,b).
  3. Calculate the average rate of change using the formula f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}.
  4. Conclude that there exists a point c∈(a,b)c \in (a,b) where the derivative f′(c)f'(c) equals this average rate.
  5. Note that the point cc is not necessarily unique.

Example

For a function satisfying the hypotheses, the theorem guarantees a tangent line at some interior point cc that is parallel to the secant line connecting the endpoints (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)).

Common misconceptions

  • Believing that the point cc is always unique.
  • Thinking that the theorem applies if the function is only continuous on (a,b)(a,b) but not at the endpoints.
  • Assuming the conclusion holds for functions that are differentiable on [a,b][a,b] but not continuous (which is impossible, but learners might confuse the domains).

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