Connect average change to an instantaneous slope

A secant line joins two points on a graph. Its slope records the average change across the whole interval. A tangent records the instantaneous rate at one point. The Mean Value Theorem connects these two measurements: for a function continuous on [a, b] and differentiable on (a, b), with a < b, at least one interior point has the same tangent slope as the secant.

The conclusion concerns a rate of change, rather than the average height of the graph. If a graph describes position over time, the secant gives average velocity. The theorem then says that some instantaneous velocity matches that average, provided the position function meets the assumptions.

Take f(x) = x² on [0, 2]. The endpoints are (0, 0) and (2, 4), so the secant slope is 2. The derivative is f′(x) = 2x. A matching tangent therefore occurs at c = 1, where f(1) = 1. The secant and tangent are parallel, but they are different lines.

Before moving a control, predict the result on [−2, 2]. Both endpoints now have height 4, so the secant is horizontal. The matching tangent is at c = 0 because f′(0) = 0. Symmetry explains this example, but symmetry is not an assumption of the theorem.

1. Open the Mean Value Theorem module and select f(x) = x².
2. Set the interval endpoints to 0 and 2. Compare the displayed secant slope and interior matching point with the calculation above.
3. Move the left endpoint to −2 while keeping the right endpoint at 2. Explain why the slope becomes zero before checking the visualization.
4. Select f(x) = x³/3 − x on [−2, 2]. Look for two matching tangents rather than assuming a unique answer.

For the cubic example, f(2) = 2/3 and f(−2) = −2/3, giving a secant slope of 1/3. Solving c² − 1 = 1/3 yields c = ±2/√3, approximately ±1.1547. Both points lie inside the interval. Riano's matching-point search is numerical, so compare approximate values rather than expecting every displayed digit to be exact.

Continuity and differentiability matter. On paper, consider f(x) = |x| on [−1, 1]. Its secant slope is zero, but the derivative is −1 on the left and +1 on the right. At the corner x = 0, the derivative is undefined. There is no interior tangent of slope zero. This is not a counterexample to the theorem: the differentiability assumption fails. The module's available smooth functions are not an experiment with this corner function.

The interval must also have positive length. When a = b, the secant formula divides by zero. The module maintains a small positive interval; this safeguard is a control constraint, not an extension of the theorem to a zero-length interval. A visual match in one experiment illustrates a theorem but does not prove it for all functions.

For f(x) = x² on [1, 3], calculate the secant slope and matching point without the module. The endpoint heights are 1 and 9, so the slope is (9 − 1)/(3 − 1) = 4. Since 2c = 4, the point is c = 2. Now use the visualization to check your prediction.

Read the companion theory article for the formal statement. For a further explanation of the theorem and its assumptions, see MIT OpenCourseWare: Mean Value Theorem.