Langrange's Mean Value Theorem

Theorem 7:

Let f be real valued function in [a,b] such that,

• f is continuous in [a,b].

• f is differentiable in (a,b).

Geometrical meaning

Let A (a,f (a)) and B (b,f (b)) be two points on f (x), then

Note:

The value of c obtained need not be unique.

Working Rule to Verify Lagrange's Mean Value Theorem

(Mean Value Theorem)

Step 1:

Show the function f (x) is continuous on the closed interval [a, b].

Step 2:

Find f '(x) and examine if it is defined at every point on the open interval (a, b). If f '(x) is defined for all x (a, b), then the function is differentiable.

Step 3:

If the above condition are satisfied, then Mean Value Theorem is applicable.

Step 4:

If Mean value theorem is applicable, solve the equation

Show that one of the roots lie in the open interval (a, b). This verifies the Mean Value Theorem.

Example:

Verify mean value theorem for the function

f (x) = (x - 4) (x - 6) (x - 8) in [4,10]

Step1:

We know that every polynomial function is continuous and product of continues functions are continuous. f (x), being product of polynomials of degree 1, is a continuous function in [4,10].

Step 2:

f ' (x) = (x - 6) (x - 4) + (x - 4) (x - 8) + (x - 6) (x - 8)

f '(x)= (x² -10x + 24) + (x² - 12x + 32)+ (x² - 14x + 48)

= 3x² - 36x + 104

f '(x) is defined for all values on the interval (4,10).

\ f '(x) is differentiable.

Step 3:

Since both the condition are satisfied, Mean Value Theorem is applicable.

\ There exist c (4, 10) such that

\ f (4) = 0

f '(c) = 3c² - 36c + 104

Substituting these values in (1), we have

Since 8 (4,10), Mean Value Theorem is satisfied.