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# Mean Value Theorem

Top
 Sub Topics Calculus is a method of calculation based on the concept of limits. A function is said to be continuous over an interval if it is continuous on every point in that interval. In other words, a function is continuous in an interval if there is no break in the curve of the function in that interval. A function is differentiable in a given interval if for each point there exists a derivative of the function. A function which is continuous need not be differentiable but the function which is differentiable is always continuous. The Rolle's and Mean-Value theorems tells about the value of the derivative of a function at a given point, and is applied to find the maximum and minimum values of a given function.

## Definition

For a given function f(x), which is continuous in the interval [a, b] and differentiable in the interval (a, b), which has values f(a) and f(b) at a, b respectively there is a number c such that a < c < b, and

$f'(c)$ =$\frac{f(b)-f(a)}{b-a}$

This implies that there is a tangent line parallel to the line passing through (a, f(a)) and (b, f(b)).

## Proof

The equation of line passing through (a, f(a)) and (b, f(b)) will be:

$y - f(a)$ = $\frac{f(b)-f(a)}{b-a}(x-a)$ $\rightarrow$ $y$ = $\frac{f(b)-f(a)}{b-a}$ $(x-a)+ f(a)$

Let, $g(x)$ = $f(x)$-[$\frac{f(b)-f(a)}{b-a}$ $(x-a)+ f(a)$] where g(x) is continuous on [a, b] and differentiable on (a, b), g(a) = g(b) = 0. Then from Rolle's theorem we have a point c such that g'(c) = 0.

We get, $g'(c) = f'(c)$ -[$\frac{f(b)-f(a)}{b-a}$] = 0 $\rightarrow$ $f'(c)$ =[$\frac{f(b)-f(a)}{b-a}$]

## Mean-Value Theorem for Integrals

The mean-value theorem for the integrals is a variation of the mean-value theorem which states that a continuous function has at least one such point where the function equals the average value of the function. If a function f(x) is continuous over [a, b] then there exists a value α in [a, b] such that, f(α) = $\frac{1}{b-a}$ $\int_{a}^{b}f(x)dx$ .

## Mean-Value Theorem for Derivatives

For a given function f(x), which is continuous in the interval [a, b] and differentiable in the interval (a, b), which has values f(a) and f(b) at a, b respectively there is a number c such that a, c, b, and

$f'(c)$ =$\frac{f(b)-f(a)}{b-a}$

Let us try and understand this by an example.

Take $f(x)$ = $x^3$ + x - 1 on the interval [1, 3]. Then there will be a point c in [1, 3] for which $f'(c)$ =$\frac{f(b)-f(a)}{b-a}$. Let us find out this point c. We have got a = 1, b = 3.

f(a) = 1 + 1 - 1 = 1

f(b) = 27 + 3 - 1 = 29

Hence, f'(c) = $\frac{27 - 1}{3 - 1}$ $\frac{26}{2}$=13 ........ (1)

To get the equation of f'(c) we first need to find f'(x).

f'(x) = 3$x^2$ + 1. Putting x = c, we get f(c) = 3$c^2$+ 1 ... (2)

Equating 1 and 2 we get,

3$x^2$ + 1 = 13 $\rightarrow$  3$c^2$ = 12 $\rightarrow$  4$c^2$= 4 $\rightarrow$ c= $\pm$ 2.

Therefore, there are two points -2, +2 but only +2 $\epsilon$ [1, 3]. So, c = 2.

## Application

The mean-value theorem is used to find if the function is increasing or decreasing in the given interval [a, b].

Let f(x) be a function such that it is continuous in [a, b] and derivable in (a, b), then

1) if f'(x) = 0, x $\epsilon$ (a, b) then f(x) is constant in [a, b].

2) if f'(x) > 0, x $\epsilon$ (a, b) then f(x) is strictly decreasing in [a, b].

3) if f'(x) = 0, x $\epsilon$ (a, b) then f(x) is strictly increasing in [a, b].

What will happen when f(a) = f(b)?

Then, $f'(c)$ = $\frac{f(b)-f(a)}{b-a}$ $\rightarrow$  $f'(c)$ = $\frac{0}{b-a}$ $\rightarrow$  f'(c) = 0.

This is the Rolle's theorem.

## Examples

Example 1: Find the value of a at which a tangent can be drawn parallel to the lines joining the end points of the curve (x + 1)$^1$in the interval [-1, 1].

Solution: f(1) = (1 + 1)$^3$ = 8; f(-1) = (-1 + 1) = 0

According to mean-value theorem,$f'(c)$ = $\frac{8 -0}{1-(-1)}$  f'(c) = 4

After differentiating we get, f'(x) = 3(x + 1)$^2$.

Hence, f'(c) = 3(c + 1)$^2$ = 4 ⇒ (c + 1 = $\frac{4}{3}$ $\Rightarrow$ (c +1)=$\pm$ $\frac{2} {\sqrt{3}}$
$\Rightarrow$  c=$\pm$ ($\frac{2}{\sqrt{3}}$) - 1

Hence, c = $\frac{2}{\sqrt{3}}$ -1 $\epsilon$ [-1,1].

Example 2: Does the mean-value theorem apply to f(y) = 3$y^2$ + 1 on [0, 1]?

Solution: f(0) = 3.0 + 1 = 1; f(1) = 3.1 + 1 = 4

According to mean-value theorem we will have a point c in [0, 1] such that,

f'(c) = $\frac{4 - 1}{1-0}$ = 3 .

Now we have, f'(y) = 6y.

f'(c) = 6c = 3

We have, c = 0.5 which belongs to [0,1]. Hence, the mean-value theorem applies to f(y) = 3$y^2$ + 1 on [0, 1].