The Mean Value Theorem
Over an interval where a function is continuous and differentiable, some interior point has instantaneous rate equal to the average rate across the whole interval. Most facts linking a derivative to a function's behaviour rest on this.
What a learner can do afterwards
- State the hypotheses and find the guaranteed point for a given function
- Give a function that fails the theorem and identify the hypothesis it breaks
- Use the theorem to argue that a zero derivative on an interval forces a constant function
1 · Read
Take a car that covers 90 km in one hour. Its average speed is 90 km per hour, even if it sped up and slowed down along the way. The mean value theorem says the speedometer showed exactly 90 at some instant. In symbols, the average rate from a to b is f of b minus f of a, over b minus a. The theorem promises a point c inside where f prime of c equals that average.
The promise needs two conditions. The function must be continuous on the closed interval from a to b, and differentiable on the open interval between them. Endpoints only need continuity, since a derivative wants slopes from both sides. Picture the secant line joining the endpoints: the theorem gives a tangent line parallel to it, touching the curve somewhere inside. When the endpoint values match, the secant is flat, so some c has f prime of c equals 0. For g of x equals x squared minus 4 x plus 3 on 1 to 3, both values are 0 and c equals 2.
Take f of x equals 2 x squared on 1 to 3. First the average rate: f of 1 is 2 and f of 3 is 18, so 18 minus 2 over 3 minus 1 is 8. The derivative is 4 x, so set 4 c equals 8 and get c equals 2, which sits inside. A polynomial passes everywhere, so both conditions hold with no extra work. A second example: f of x equals x minus 3 over x on 1 to 3. The average rate is 2, and f prime of x is 1 plus 3 over x squared. Setting that equal to 2 gives c squared equals 3, so c is root 3, about 1.73, inside the interval.
When a condition fails, the promise evaporates. The absolute value of x minus 2 has a sharp corner at x equals 2, where the slope jumps from negative 1 to 1, so no derivative exists there. And 1 over x blows up at x equals 0, so it is not continuous on any closed interval holding 0. Failing the conditions means no guarantee, though the conclusion may still hold by luck.
The theorem also bounds growth and proves constancy. If f of 1 is 2 and the derivative stays at most 5, then f of 4 is at most 2 plus 3 times 5, which is 17. And if the derivative is 0 at every interior point, then f of b minus f of a equals 0 times b minus a, so every value matches and the function is constant.
A smooth curve always has an interior point whose tangent matches its secant.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.