Mean Value Theorem

History of Mean Value Theorem

Mean Value Theorem was first defined by Vatasseri Parameshvara Nambudiri (a famous Indian mathematician and astronomer), from the Kerala school of astronomy and mathematics in India in the modern form, it was proved by Cauchy in 1823.

Its special form of theorem was proved by Michel Rolle in 1691; hence it was named as Rolle’s Theorem. It was Moritz Wilhelm Drobisch of Germany who used this name first in 1834 and then Giusto Bellavitis of Italy in 1846.
 

What is Mean Value Theorem?

The Mean Value Theorem states that for any given curve between two endpoints, there must be a point at which the slope of the tangent to the curve is same as the slope of the secant through its endpoints.

If f(x) is a function, so that f(x) is continuous on the close interval [a, b] and also differentiable on the open interval (a, b), then there is point c in (a, b) that is, a < c < b such that

The mean value theorem is also known as Lagrange’s Mean Value Theorem or first mean value theorem.
 

Graphical Interpretation of Mean Value Theorem

Here the above figure shows the graph of function f(x).

Let A = (a, f (a)) and B = (b, f (b))

At point c where the tangent passes through the curve is (c, f(c)).

The slope of the tangent line is same as secant line that is, both are parallel to each other.

Proof of the Mean Value Theorem

To prove the expression

We have to assume

h(x) = f(x) –sx                (i)

Where s is a constant.

As we know that function is continuous on [a, b] and differentiable on (a, b), so the function h(x) is also having the same properties.

Now according to Rolle’s Theorem 

By Rolle's Theorem, as function ‘h’ is {\displaystyle g}differentiable and {\displaystyle g(a)=g(b)}h (a) = h (b) so that hꞌ(c) = 0.

Hence h(c) = f(c) –sx

Example 

If a, b, are two numbers with a < b, show that a real number c can be found between a and b such the 3c² = b² + ab + a².

Solution: Consider the function f(x) = x³ 

It is continuous and differentiable in (a, b).

Hence by LMVT, there exist a point c such that  a < c < b and f’(c) = f'(c) = f(b)–f(a)/b–a

f'(x) = 3x², we get 3c² = b³–a³/b–a = b² + ba + a².

What is Rolle’s Theorem in calculus?

It is a special case of mean value theorem which says that if y = f (x) be a given function and satisfies the conditions:

(i) f (x) is continuous on the closed interval [a , b]

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

(iii) f (a) = f (b)

Then there must be one point c which comes between a and b so that a < c < b.

Hence f'(c) = 0 at least once for some x ∈ (a, b)

that is,  the first derivative is zero or you can say that the slope of the tangent line to the curve of function is zero.

Note: 

(1) There can be more than one such c.

(2) Think! The conditions of Rolle’s theorem are necessary or sufficient or both? The answer is conditions are only sufficient and necessary will be clear from the following examples:

(a) Let

Here condition (i) is violated

However f ‘(x) = 0 if x = 1/2 ∈ (0, 1)

By defining f(0) = 0 and f(1) = 3 we can see that the result is true when (i) and (iii) are violated.

(b) Let 

Clearly (ii) does not hold in (0, 2) and even so f'(3/2) = 0 

(3) If f(x) is any polynomial then between any pair of roots of f (x) = 0 lies a root of f ’(x) = 0
 

Graphical representation of the Rolle’s Theorem

Here the graph of function y = f(x) is continuous between x = a and x = b. there could be so many tangents but at least one tangent is there which is parallel to x axis. At that point fꞌ(c) = 0.

Example 1:

If ax² + bx + c = 0, a, b, c ∈ R. Find the condition that this equation would have at least one root in (0, 1).

Solution: Let f’(x) = ax² + bx + c

Integrating both sides,

 => f(x) = ax³ / 3 + bx² / 2 + cx + d 

=> f(0) = d and f(1) = a/3 + b/2 + c + d

Since, Rolle’s theorem is applicable

=> f(0) = f(1) => d = a/3 + b/2 + c + d

=> 2a + 3b + 6c = 0

Hence required condition is 2a + 3b + 6c = 0

Example 2: 

Find the function is constant function or not? f(x) = 1/x at a = -1 and b = 1.

Solution: For function f(x) = 1/x where a = -1 and b = 1.

We have to solve the right hand side

On the left hand, for any cε (-1,1), ∈ not equal to 0, we have 

So the equation 

does not have a solution in c. This does not contradict the Mean Value Theorem, since f(x) is not even continuous on [-1, 1].

This shows that the first derivative of a constant function is 0.

Let f(x) be a differentiable function on an interval I, with f '(x) = 0, for every x ϵ I. Then for any a and b in I, the Mean Value Theorem implies 

There is some ‘c’ between a and b. So we implies our assumption

f(b) – f(a) = 0. (b – a) = 0.

Thus f (b) = f (a) for any a and b in I, which means that f(x) is constant.

Example 3:

Show that the equation 2x³ + 3x² + 6x + 1 = 0 has exactly one real root.

Solution: Let the function as f(x) = 2x³ + 3x² + 6x + 1.

Now we have 

f (0) = 2x³ + 3x² + 6x + 1

= 2(0)³ + 3(0)² + 6(0) + 1

= 1

f (-1) = 2x³ + 3x² + 6x + 1

= 2(-1)³ + 3(-1)² + 6(-1) + 1

= - 2 + 3 - 6 + 1

= - 4

Here the Mean Value Theorem shows that there is a point c between 0 and -1 so that f(c) =0.

Therefore this equation has at least one real root.

Now we will check whether this equation has one and only one real root or more than that.

For this we will assume that there are at least two roots c₁and c₂, with c₁ < c₂. Then we have f (c₁) = 0 and f (c₂) = 0.

Rolle's Theorem says that a point c between c₁ and c₂ such that

f '(c) = 6c² + 6c + 6 = 0.

But the quadratic equation 6c² + 6c + 6 = 0 does not have real roots, resulting a flaw to our assumption that f(x) had at least two roots.

Hence this equation has only one real root.

What is the difference between the mean value theorem and the Rolle’s Theorem?

More Readings

Mean Value Theorem