Subject: Calculus

# Mean Value Theorem

Imagine yourself filling up a container with water from a faucet. Let's say that the container can hold water up to at most 5 liters. Now, as you turn on the faucet, you notice that it takes 5 minutes for the faucet to finally fill up the container. And in moment's time, you easily calculated that the average rate of the water flow through the faucet is obviously 5 liters every 5 minutes. So the rate would be 1 liter per minute. But you know in yourself that during the time you watched the scenario, sometimes the faucet flows out water very fast and sometimes you noticed the flow to be very slow, especially when you had just turn on the faucet. But you don't know exactly the flow rate at that moment and the average speed says that it must be 1 liter per minute.

So the question is, is it possible that during the time interval of 5 minutes, the water flow rate had somehow attain a rate of 1 liter per minute? Why?

Well, the answer is obviously yes and the reason is the **mean value theorem**. The mean value theorem tells us that at some point during the flow of the water, the flow rate was exactly 1 liter per minute; which is equal to its average speed. To understand why it said so, let us go to the mathematical definition of the mean value theorem.

## Mean Value Theorem

Let f be a function very much defined on a Euclidean plane such that

(i) it is continuous on the closed interval [a, b], and

(ii) is differentiable on the open interval (a, b).

Then there is a number on the open interval (a, b) such that f'(c) = \frac{f(b) - f(a)}{b-a}.

Geometrically, if we draw a line connecting the endpoints of the interval (a, b) then the expression f'(c) = \frac{f(b) - f(a)}{b-a} is the slope of that line. This slope just represents the average derivative of the function on the said interval. The theorem also states that at some point in the interval, the tangent line L_{t} at that point is parallel to the line connecting the endpoints of the interval.

### Special case:

When f(b) = f(a), then f'(c) =0. This case is called as the **Rolle's Theorem** which states that there must be a point in the curve such that the tangent line L_{t} to the curve at that point (c) is a horizontal line.

Example, you have a function f(x) = \frac{2x+3}{3x-2}, let us find all the number c between the interval (a, b) such that f'(c) = \frac{f(b) - f(a)}{b-a}.

Now, f(a) = f(1) = 5 and f(b) = f(5) = 1. Thus, \frac{f(b) - f(a)}{b-a} = \frac{1-5}{4} = -1. Therefore, f'(c) = -1. On the other hand, let's get the first derivative of the function f. That is,

This leads to

but we know that f'(c)=-1, so

In which we algebraically solve for c to get

But we know that the interval is from points 1 to 5 and obviously the negative value of c is not on the interval, so c is given by

### Example #1

Which theorem accommodates the case when the derivative of f at point c is equal to zero?

Roll's Theorem

### Example #2

A parabola has its vertex on the origin

\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy = 0

### Example #3

What happens if the function f under the integral sign is independent of either x or y?

\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy = 0

### Example #4

What happens if the function f under the integral sign is independent of either x or y?

\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy = 0

### Example #5

What happens if the function f under the integral sign is independent of either x or y?

**NEX TOPIC**: Logarithmic derivative