The Mean Value Theorem (MVT) for derivatives states that if the following two statements are true:
- A function is a continuous function on a closed interval [a,b], and
- If the function is differentiable on the open interval (a,b),
…then there is a number c in (a,b) such that:
The Mean Value Theorem is an extension of the Intermediate Value Theorem.
The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.
There is also a mean value theorem for integrals.
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The Common Sense Explanation
The “mean” in mean value theorem refers to the average rate of change of the function. It’s basic idea is: given a set of values in a set range, one of those points will equal the average. This is best explained with a specific example.
Let’s say you travel from your house to work, varying your speed between 40 and 50 mph. The speedometer needle will fluctuate between 40 and 50, and let’s say you average 45 mph. As the needle moves from 40 to 50, it has to pass this point at least once. I picked 45 mph arbitrarily, but you could pick any number between 40 and 50 (i.e. the “closed interval”) and the needle would have to pass that point.
Mean Value Theorem Example Problem
Example problem: Find a value of c for f(x) = 1 + 3√(x – 1) on the interval [2,9] that satisfies the mean value theorem.
Step 2: Place your answer from Step 1 into the formula in the f′ (c) position:
Step 3: Plug in the two boundaries (from the question: 2 and 9) into the formula. “b” is the highest value on the number line, and “a” is the smallest value. Make sure you put those values in the numerator and the denominator:
Step 4: Work the right side of the equation. For f(2) and f(9), you solve by plugging each value, 2 and 9, into the formula from the question.
In other words, solve f(x) = 1 + 3√(2 – 1) and f(x) = 1 + 3√(9 – 1). I used Google’s calculator, which solves cubed roots if you type in the words (i.e. 1 + the cubed root of (2 – 1)).
- Multiply by 3 and rewrite as 2⁄3 power.
- Multiply by (x – 1)2⁄3.
- Multiply by 7⁄3.
- Raise to 3⁄2.
- Add 1.
Note: If you have two solutions (e.g. a positive and negative value), check your interval to make sure both values are in the interval. In some cases, only one of the values will be in the interval, giving you just one solution.
Rolle’s theorem is a special case of the mean value theorem. It states that for any continuous, differentiable function with two equal values at two distinct points, the function must have a point where the first derivative is zero.
Theorem in Graphical Terms
- Take any interval on the x-axis (for example, -10 to 10). Make sure two of your function values are equal.
- Draw a line from the beginning of the interval to the end. It doesn’t matter if the line is curved, straight or a squiggle—somewhere along that line you’re going to have a horizontal tangent line where the derivative, (f′) is zero. Try it!
Rolle’s Theorem in Math Terms
The standard version of Rolle’s Theorem goes like this: Let’s say you have a function f with the following characteristics:
- It’s differentiable on the open interval (a,b),
- It is a continuous function on the closed interval [a,b],
- f(a) = f(b).
Then there is some c, with a ≤ c ≤ b such that f′(c) = 0.
This is illustrated by the diagram below, which shows a real-valued, continuous function on a closed interval. According to our theorem, the function equals itself at the two endpoints of the interval means that there is a point where the tangent line is horizontal.
Note that Rolle’s lemma tells us that there is a point with a derivative of zero, but it doesn’t tell us where it is. It doesn’t give us a method of finding that point either. Still, this theorem is important in calculus because it is used to prove the mean-value theorem.
How to use Rolle’s Theorem
Example question: Use Rolle’s theorem for the following function:
f(x) = x2 – 5x + 4 for x-values [1, 4]
Step 1: Find out if the function is continuous. You can only use Rolle’s theorem for continuous functions.
Step 2: Figure out if the function is differentiable. If it isn’t differentiable, you can’t use Rolle’s theorem. the easiest way to figure out if the function is differentiable is to simply take the derivative. If you can take the derivative, then it’s differentiable.
f′(x) = 2x – 5
Step 3: Check that the derivative is continuous, using the same rules you used for Step 1.
f′(x) = 2x – 5 is a continuous function.
If the derivative function isn’t continuous, you can’t use Rolle’s theorem.
Step 4: Plug the given x-values into the given formula to check that the two points are the same height (if they aren’t, then Rolle’s does not apply).
- f(1) = 12 -5(1) + 4 = 0
- f(4) = 42 -5(4) + 4 = 0
Both points f(1) and f(4) are the same height, so Rolle’s applies.
Step 5: Set the first derivative formula (from Step 2) to zero in order to find out where the function’s slope is zero.
- 0 = 2x – 5
- 5 = 2x
- x = 2.5
The function’s slope is zero at x = 2.5.
Rolle’s theorem has a long history: we have reason to believe it was known by Indian mathematician Bhaskara II, who lived between 1114-1185. It is named after Michel Rolle, who published a proof of the polynomial case in 1691. The name was first used in 1834, by mathematician and philosopher Moritz Wilhelm Drobisch.
Ghosh, J. (2004). How to Learn Calculus of One Variable. New Age International.
Hosch, Wiliam L. Rolle’s. Encyclopædia Britannica
Publisher: Encyclopædia Britannica, inc. Date Published: August 05, 2011
Retrieved from https://www.britannica.com/science/Rolles-theorem
on April 04, 2019
Stephanie Glen. "Mean Value Theorem & Rolle’s Theorem" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/calculus-problem-solving/intermediate-value-theorem/mean-value-theorem/
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