Lagrange's Mean Value Theorem

Introduction to Lagrange’s Mean Value Theorem:

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (differentiable) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.

(Source – Wikipedia)

Lagrange’s Mean Value Theorem – Proof:

The followings statements are the proof for the Lagrange’s Mean Value Theorem.

If a function f(x) is derivable in the interval [a, b], then, there exists at least one value ‘c’ of x lying within [a, b] such that `(f(b)-f(a))/(b-a)` = f’(c).

Proof:

To prove the theorem, we defined a new function `phi` (x) involving f(x), which is designed so as to satisfy the condition of Rolle’s Theorem.

Let `phi` (x) = f(x) + A x where A is a constant to be determined such that `phi` (a) = `phi` (b).

Thus, f(a) + Aa = f(b) + Ab

Therefore, A = `(-f(b)-f(a))/(b-a)`

Now, f(x) is derivable in [a, b]. Also x is derivable and A is a constant. Therefore, `phi` (x) is derivable in [a, b] and its derivative is f’(x) + A.

Thus, `phi` (x) satisfies all the conditions of Rolle’s Theorem.

There is, therefore, at least one value ‘c’ of x, lying within (a, b) such that `phi` ’(c) = 0

Therefore, 0 = `phi` ’(c) = f’(c) + A, that is –A = f’(c)

Or

` (-f(b)-f(a))/(b-a)` = -f’(c)

Or

`(f(b)-f(b))/(b-a)` = f’(c) →(i)

In certain books Lagrange’s Mean Value Theorem is stated as follows.

If a function f(x) defined in the closed interval [a, b] is such that

(i) f(x) is closed interval in continuous as [a, b], and

(ii) f(x) is derivable in the open interval (a, b), then there exists at least one value ‘c’ of x within the interval [a, b] such that f’(c) = `(f(b)-f(a))/(b-a)` .

This seems to be unnecessary as every finitely derivable is continuous and therefore the definition given by us is quite sufficient. I have recently faced lot of problem while learning define rational number, But thank to online resources of math which helped me to learn myself easily on net.

Hence we proved the Lagrange's Mean Value Theorem.

Lagrange’s Mean Value Theorem – Example Problem:

The followings examples are the proof for the Lagrange’s Mean Value Theorem.

Verify Lagrange’s Mean Value Theorem for the f(x) = 2x – x^2 in the interval [0, 1].

Solution:

The function 2x – x^2, being a polynomial is derivable in [0, 1].

f(x) = 2x – x^2 also f’(x) = 2 – 2x.

By Lagrange’s Mean Value Theorem, there exists c in (0, 1) such that

` (f(1)-f(0))/(1-0)` = f’(c)

Or

f(1) – f(0 = f’(c)

Now f(1) =` 2 * 1` – 12 = 1, f(0) = ` 2 * 0` – 0, f’(c) = 2 – 2c

1 – 0 = 2 – 2c `rArr` c = ` 1/2` which lies in(0, 1).

Hence we can verify Lagrange’s Mean Value Theorem.