Mean Value Theorem
In short: Mean value theorem says that a smooth, unbroken curve over an interval has at least one point where the instantaneous slope equals the average slope across the whole interval.
AP Calculus ABCalculusAP Calculus AB Unit 5
What is Mean Value Theorem?
The mean value theorem connects a trip’s average speed to its speedometer. If a car covers 66 miles in exactly one hour and never teleports, at some instant the speedometer must read exactly 66 mph.
Formally, let \(f\) be continuous on \([a, b]\) and differentiable on \((a, b)\). Then some \(c\) in \((a, b)\) satisfies
\[f^{\prime}(c) = \frac{f(b) – f(a)}{b – a}\]
Geometrically, there is a tangent line parallel to the secant line joining the endpoints.
How to spot it
- A problem gives an average rate on an interval and asks you to prove the instantaneous rate matches it somewhere.
- The prompt says “find all values c guaranteed by the theorem” for a named function and interval.
- A speed, temperature or growth story needs a justification that a rate reached an average value.
- You must check whether a function is continuous and differentiable on an interval before concluding.
The method
- Verify the two hypotheses: continuity on \([a, b]\) and differentiability on \((a, b)\).
- Compute the average rate of change \(\frac{f(b) – f(a)}{b – a}\).
- Solve \(f^{\prime}(c)\) equal to that number for \(c\).
- Keep only solutions strictly between \(a\) and \(b\), and cite the mean value theorem in your answer.
Mean Value Theorem: solved example
Problem
For \(f(x) = x^3 – 3x\) on \([0, 2]\), find the value of \(c\) guaranteed by the mean value theorem.
Solution
A polynomial is continuous and differentiable everywhere, so the hypotheses hold. Compute \(f(0) = 0\) and \(f(2) = 8 – 6 = 2\), so the average rate is \(\frac{2 – 0}{2 – 0} = 1\).
Solve \(f^{\prime}(c) = 3c^2 – 3 = 1\), giving \(c^2 = \frac{4}{3}\), so \(c = \pm\frac{2}{\sqrt{3}}\).
Only the positive root lies in \((0, 2)\): \(c = \frac{2}{\sqrt{3}} \approx 1.155\).
Try it
Question 1
A truck passes mile marker 12 at 1:00 p.m. and mile marker 78 at 2:00 p.m. What does the theorem say about its speed?
Show the answer
The average speed is \(\frac{78 – 12}{1} = 66\) mph, so at some moment between 1:00 and 2:00 the truck was going exactly 66 mph.
Question 2
Does the theorem apply to \(f(x) = |x|\) on \([-2, 3]\)?
Show the answer
No. The function is not differentiable at \(x = 0\), inside the interval, so a required condition fails.
Mean Value Theorem vs. Intermediate Value Theorem
The intermediate value theorem needs only continuity and promises that a chosen output height is reached. This theorem needs differentiability too and promises a point where the slope equals the average slope. One is about heights of the graph, the other about steepness.
Keep going
Lessons
Worksheets and quizzes
- Mean Value Theorem and Extreme Value Theorem – Worksheet
- Mean Value Theorem and Extreme Value Theorem – Quiz
Related words
- Intermediate Value Theorem
- Extreme Value Theorem
- Derivative
- Equation of a Tangent Line
- Critical Point
Whole dictionary: A–Z · AP Calculus AB · Calculus