Key takeaway: The mean value theorem says a smooth, unbroken curve between two endpoints has some interior point where the tangent slope exactly equals the average slope from one endpoint to the other.
Grade 12 – AP Calculus ABAP Calculus AB Unit 5Calculus
Definition of Mean Value Theorem
The mean value theorem (MVT) ties average change to instantaneous change. If a smooth curve runs between two points, then at some moment along the way the instantaneous rate must equal the average rate for the whole stretch.
A cyclist who rides 24 miles in 1.5 hours averages 16 miles per hour, so the speedometer read exactly 16 at least once. On a graph, some tangent line is parallel to the secant line joining the endpoints.
Formula and notation
Take a function \(p\) that is continuous on the closed interval from \(m\) to \(n\) and differentiable on the open interval between them. Some \(t\) between \(m\) and \(n\) satisfies
\[ p^{\prime}(t)=\frac{p(n)-p(m)}{n-m}. \]
The right side is the average rate of change, while \(p^{\prime}(t)\) is the instantaneous rate at \(t\).
Key facts
- Both hypotheses matter: no breaks on the closed interval, and no corners or cusps inside it.
- The mean value theorem promises at least one such point but gives no method to find it.
- When f(a) equals f(b), the conclusion is f'(c) = 0, known as Rolle’s theorem.
- A function whose derivative is zero everywhere on an interval must be constant there.
Mean Value Theorem example
Find the number \(c\) promised by the theorem for \(f(x)=\sqrt{x}\) on \([1,9]\).
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The square root is continuous on \([1,9]\) and differentiable on \((1,9)\). Average slope: \(\frac{3-1}{9-1}=\frac{1}{4}\).
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The derivative is \(f^{\prime}(x)=\frac{1}{2\sqrt{x}}\).
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Set \(\frac{1}{2\sqrt{c}}=\frac{1}{4}\), so \(\sqrt{c}=2\).
The tangent slope matches the average slope at \(c=4\), which lies inside \((1,9)\).
Common mistake
Watch out
A student applies the theorem to \(f(x)=|x|\) on \([-1,4]\), where the average slope is \(\frac{4-1}{5}=0.6\), and hunts for a \(c\) with \(f^{\prime}(c)=0.6\).
Why it is wrong
No such \(c\) exists, because the slope of \(|x|\) is only \(-1\) or \(1\). The theorem does not apply: \(f\) is not differentiable at \(x=0\), which lies inside the interval. Verify both hypotheses before trusting the mean value theorem’s conclusion.
Related terms and lessons
- Derivative
- Intermediate Value Theorem
- Extreme Value Theorem
- Average Rate of Change
- Equation of a Tangent Line
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