# Mean Value Theorem Proof

Mean value theorem is one of the most useful tools in both differential and integral calculus. It has very important consequences in differential calculus and helps us to understand the identical behavior of different functions.

The hypothesis and conclusion of the mean value theorem shows some similarities to those of Intermediate value theorem. Mean value theorem is also known as Lagrange’s Mean Value Theorem. This theorem is abbreviated as MVT.

## Mean Value Theorem Statement

Suppose f(x) is a function that satisfies below conditions:

1. f(x) is Continuous in [a,b]
2. f(x) is Differentiable in (a,b)

Then, there exists a number c, s.t. a < c < b and

f(b) – f(a) = f ‘(c) (b – a)

 Special Case: When f(a) = f(b). Then, there exists at least one c with a < c < b such that f'(c) = 0. This case is known as Rolle’s Theorem.

## Proof of Mean Value Theorem

The Mean value theorem can be proved considering the function h(x) = f(x) – g(x) where g(x) is the function representing the secant line AB. Rolle’s theorem can be applied to the continuous function h(x) and proved that a point c in (a, b) exists such that h'(c) = 0. This equation will result in the conclusion of mean value theorem.

Consider a line passing through the points, (a, f(a)) and (b, f(b)). Equation of line is

y – f(a) = {f(b) – f(a)}/(b-a) . (x – a)

or y = f(a)+ {f(b) – f(a)}/(b-a) . (x – a)

Let h be a function define difference between any function f and the above line.

h(x) = f(x) – f(a) – {f(b)-f(a)}/(b-a) . (x – a)

using “Rolle’s theorem”, we have

h'(x) = f'(x) – {f(b)-f(a)}/(b-a)

Or f(b) – f(a) = f'(x) (b – a). Hence Proved.

### Physical Interpretation of Mean Value Theorem

Since (f(b)−f(c))/(b−a) is the average change in the function over [a, b], and f'(c) is the instantaneous change at ‘c’, the mean value theorem states that at some interior point the instantaneous change is equal to average change of the function over the interval.

### Corollaries of Mean Value Theorem

Corollary 1: If f(x) = 0 at each point of x of an open interval (a, b), then f(x) = C for all x in (a, b) where C is a constant.

Corollary 2: If f'(x) = g'(x) at each point x in an open interval (a, b), then there exists a constant C such that f(x) = g(x) + C.

The first corollary confirms that if the derivative of a function is zero then the function is a constant function. The second corollary says that the graphs of functions with identical derivatives differ only by a vertical shift. This property is used in solving initial value problems in integral calculus.

## Application of Mean Value Theorem

Mean value theorem is the relationship between the derivative of a function and increasing or decreasing nature of function. It basically defines the derivative of a differential and continuous function. Below are few important results used in mean value theorem.

1. Let the function be f such that it is, continuous in interval [a,b] and differentiable on interval (a,b), then

f'(x) = 0, x ∈ (a,b), then f(x) is constant in [a,b].

2. Let f and g be a functions such that, f and g are continuous in interval [a,b] and differentiable on interval (a,b),

f'(x) = g'(x), x ∈ (a,b), thenf(x) – g(x) is constant in [a,b]

3. Strictly Increasing Function

Let the function be f such that, continuous in interval [a, b] and differentiable in interval(a,b)

f'(x) > 0, x ∈ (a,b), then f(x) is strictly increasing function in [a,b]

4. Strictly Decreasing Function

Let the function be f such that, continuous in interval [a,b] and differentiable in interval (a, b)

f'(x) < 0, x ∈ (a,b), then f(x) is strictly decreasing function in [a,b].

### Cauchy Mean Value Theorem

 Statement: Let f and g be functions defined on [a,b] such both are continuous in closed interval [a,b] both are derivable in open interval (a,b) g'(x) ≠ 0 for any x ∈ (a,b) then there exists at least one point c ∈ (a,b) such that $\frac{f'(c)}{g'(c)} = \frac{f(b) – f(a)}{g(b) – g(a)}$

If we take g(x) = x for every x ∈ {a,b] in Cauchy’s mean value theorem, we get

$\frac{f(b) – f(a)}{b – a}$ = f'(c) which is Langrange’s mean value theorem. This is also called an extended mean value theorem.

### Generalized Mean Value Theorem

If we have three functions f, g and h defined in such that,

f, g and h are continuous in [a,b],

f, g and h are derivable in (a,b).

Then, there exists a real number c ∈ (a,b) such that,

$\begin{vmatrix} f'(c) & g'(c) & h'(c) \\ f(a) & g(a) & h(a) \\ f(b) & g(b) & h(b) \end{vmatrix} = 0$

In the above, if we take g(x) = x and h(x) = 1, we obtain Langrange’s mean value theorem and if we take h(x) = 1, then we obtain Cauchy’s mean value theorem.

## Mean Value Theorem for Derivatives

With the help of mean value theorem, we approximate the derivative of any function. Theorem can build a relationship between the slope of a tangent line and the secant line on a curve.

If f is differentiable over (a,b) and continuous over [a,b] then there exists a point

c in such a way that f′(c) = {f(b)−f(a)}/(b−a)

It shows that the actual slope is equal to the average slope at some point in the closed interval. Geometrically, we can say that between two end points of the curve, we have at least one point on the curve where the slope of the tangent line equal to the slope of the secant line passing through A and B.

Theorems for differentiation

Limits continuity and differentiability

## Mean Value Theorem Examples

Given below are some of the examples of mean value theorem for better understanding.

Question 1: Find the value or values of c, which satisfy the equation $\frac{f(b) – f(a)}{b – c}$ = f'(c) as stated in Mean Value theorem for the function f(x) = $\sqrt{(x – 1)}$ in the interval [1, 3].

Solution:

First the conditions of Mean value theorem are to be checked.

f(x) is continuous in its Domain [0, $\infty$) and hence in the given interval [1, 3].

f(x) is also differentiable in the given interval. Plugging a = 1 and b = 3 in the expression on the left side of the equation,

$\frac{f(3)- f(1)}{3 – 1}=\frac{\sqrt{2} – 0}{2}\frac{\sqrt{2}}{2}$ .

Now, the derivative of the function can be found using chain rule as

f'(x) = $\frac{1}{2 \sqrt{x – 1}}$ .

Hence, the equation can be formed as

f'(c ) = $\frac{1}{2 \sqrt{x – 1}}$ = $\frac{\sqrt{2}}{2}$

Cross multiplying and squaring the equation reduces to,

2(c – 1) = 1 which gives the solution as c = 3/2 which lies in the given interval [1, 3].

Question 2: Verify Rolle’s theorem for the function f(x) = x2 – 8x + 12 on (2, 6)

Solution:

Since a polynomial function is continuous and differentiable everywhere, f(x) is differentiable and continuous conditions of Rolle’s theorem is satisfied.

f (2) = 22 – 8 (2) + 12 = 0

f (6) = 62 – 8(6) + 12 = 0

This implies, f(2) = f(3)

Therefore, Rolle’s theorem is applicable for the given function f(x).

There must exist c ∈ (2, 6) such that f'(c) = 0

f'(x) = 2x – 8

f'(c) = 2c – 8

2c – 8 = 0

c = 4 ∈ (2,6)

Therefore, Rolle’s theorem is verified.

Question 3: For the function $f(x)={{e}^{x}},a=0,b=1$, then find the value of c in the mean value theorem.

Solution:

$\frac{f(b)-f(a)}{b-a}=f'(c) \\ \frac{{{e}^{b}}-{{e}^{a}}}{b-a}=f'(c)\\\frac{e-1}{1-0}={{e}^{c}}\\\Rightarrow c=\log (e-1)\\$

Question 4: From mean value theorem, $f(b)-f(a)= (b-a)f'({{x}_{1}}); a<{{x}_{1}}, then find the value of ${x}_{1}$.

Solution:

$f'({{x}_{1}})=\frac{-1}{x_{1}^{2}} \\ \text therefore \frac{-1}{x_{1}^{2}}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}=-\frac{1}{ab}\\\Rightarrow {{x}_{1}}=\sqrt{ab}\\$

Question 5: Let f(x) satisfy all the conditions of mean value theorem in [0, 2]. If f (0) = 0 and $|f'(x)|\,\le \frac{1}{2}$ for all x, in [0, 2] then

$A)f(x)\le 2\\ B)|f(x)|\le 1\\ C)f(x)=2x\\ D)f(x)=3 \text for \ at \ least \ one \ x \ in \ [0, 2]\\$

Solution:

$\frac{f(2)-f(0)}{2-0}=f'(x)\Rightarrow \frac{f(2)-0}{2}=f'(x) \\\Rightarrow \frac{df(x)}{dx}=\frac{f(2)}{2}\Rightarrow f(x)=\frac{f(2)}{2}x+c \\\text So \ f(0)=0\Rightarrow c=0;\\f(x)=\frac{f(2)}{2}x …..\text (i)$

Given $|f'(x)|\le \frac{1}{2}\Rightarrow \left| \frac{f(2)}{2} \right|\le \frac{1}{2}..\text (ii) \\ (i) |f(x)|=\left| \frac{f(2)}{2}x \right|=\left| \frac{f(2)}{2} \right||x|\le \frac{1}{2}|x|\text (from \ (ii))$

In [0, 2], for maximum $x(x=2) |f(x)|\le \frac{1}{2}.\,\,\,2\Rightarrow |f(x)|\le 1\\$