# Mean Value Theorem: How to Use It in Easy Steps

Calculus > Mean Value Theorem

## What is the Mean Value Theorem?

For any arc between any two endpoints (a,b), a point c exists where the tangent at c is parallel to the secant line through (a,b).

The Mean Value Theorem (MVT) states that if the following two statements are true:

1. A function is continuous on a closed interval [a,b], and
2. 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.

## 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 54 mph. As the needle moves from 40 to 50, it has to pass this point at least once. I picked 54 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 1: Check that the function is continuous and differentiable. This particular function—a cubed root—is both differentiable and continuous.

Step 2: Find the derivative. Use the chain rule for this particular function.

Step 3: Plug the derivative into the left side of the formula.

Step 4: Plug the function inputs (from the question) and the function’s values into the right side of the mean value theorem formula.

Step 5: Work the right side of the equation, evaluating the function (from the question) at f(2) and f(9).

Step 6: Solve for x, using algebra:

1. Multiplying both sides by 3. Then rewriting as 23 power.
2. Multiplying by (x-1)23.
3. Multiplying both sides by 73.
4. Raising both sides to 32.
5. Adding 1 to both sides.

A negative can’t be part of the solution (considering that the interval [2,9] is positive) so the point that solves the equation is:

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