Lagrange Theorem

In this page Lagrange theorem we are going to see how to check la grange's theorem in a function.

If f (x) be a real valued function that satisfies the following conditions.

 1 f(x) is defined and continuous on the closed interval [a,b] 2 f(x) is differentiable on the open interval (a,b).

Then there exists at least one point c ∊ (a,b) such that

f ' (c) = f (b) - f (a) / (b - a)

Example 1:

Using Lagrange-theorem find the values of c.

f (x) =   x² + 4 x + 3 ,     1 ≤ x ≤ 3

Solution:

(i) f (x) is continuous on [1 ,3].

(ii) f(x) is differentiable (1,3).

f ' (x) = 2 x + 4 (1)

= 2 x + 4

f ' (c) = 2 x + 4

f(1) = 1² + 4 (1) + 3

=  1 + 4 + 3

f (1) =  8

f(3) = 3² + 4 (3) + 3

=  9 + 12 + 3

=  12 + 12

f (3) = 24

Here a = 1 and b = 3

f ' (c) = [f (b) - f (a)] / (b - a)

2x + 4  = (24 - 8)/(3 -1)

2x + 4 = 16/2

2x + 4 = 8

2x = 8 - 4

2x = 4

x = 4/2

x = 2  Lagrange theorem

Example 2:

Using Lagrange theorem find the values of c.

f (x) =   2x³ + x² - x - 1,      0 ≤ x ≤ 2

Solution:

(i) f (x) is continuous on [0 ,2].

(ii) f(x) is differentiable (0,2).

f ' (x) = 2(3x²) + 2 x - 1 - 0

= 6 x² + 2 x - 1

f (x) =   2x³ + x² - x - 1

f (0) = 2 (0)³ + (0)² - 0 - 1

f (0) = -1

f (2) = 2 (2)³ + (2)² - 2 - 1

= 2 (8) + 4 - 2 - 1

= 16 + 4 - 3

= 20 - 3

f (0) = 17

Here a = 0 and b = 2

f ' (c) = [f (b) - f (a)] / (b - a)

6 x² + 2 x - 1  = [17 - (-1)]/(0 + 2)

6 x² + 2 x - 1  = (17 +1)/2

6 x² + 2 x - 1  = 18/2

6 x² + 2 x - 1  = 9

6 x² + 2 x - 1 - 9 = 0

6 x² + 2 x - 10 = 0

3 x² +  x - 5 = 0

This equation cannot be solved using a = 3, b = 1, c = -5

x = - 1 ± √ [(1)² - 4 (3) ( -5 )]/2 (3)

x = - 1 ± √ [1 + 64]/6

x = [- 1 ± √(65)]/6

x = (- 1 ± 8.06)/6

x = (- 1 + 8.06)/6 ,  x = (- 1 - 8.06)/6

x = 7.06/6 ,  x = - 9.06/6

x = 1.17  ,  x = - 1.51

Related Topics

Lagrange Theorem to Mean Value Theorem