# Apply Lagrange theorem to the function f(x)=1/(x-1)(x-2) over the interval [3;4].

giorgiana1976 | Student

The function is continuous over the interval [3;4] (the function is discontinuously for x = 1 and x = 2). Also, the function, being continuous, it could be differentiated over the range [3 ; 4].

We'll apply Lagrange's theorem over the range [3 ; 4]:

f(4) - f(3) = f'(c)(4 - 3), if c is in the range (3 ; 4).

f'(c) = [f(4) - f(3)]/(4-3)

f(4) = 1/(4-1)(4-2)

f(4) = 1/6

f(3) = 1/(3-1)(3-2)

f(3) = 1/2

f(4) - f(3) = 1/6 - 1/2

f(4) - f(3) = (1-3)/6

f(4) - f(3) = -1/3

Now, we'll determine f'(x):

f'(x) = [1/(x-1)(x-2)]'

f'(x) = [1/(x^2 - 3x + 2)]'

f'(x) = (-2x + 3)/[(x-1)(x-2)]^2

f'(c) = (-2c + 3)/[(c-1)(c-2)]^2

But, according to Lagrange's theorem, we'll get:

f'(c) = -1/3

f'(c) = (-2c + 3)/[(c-1)(c-2)]^2

(-2c + 3)/[(c-1)(c-2)]^2 = -1/3

6c - 9 = [(c-1)(c-2)]^2

The number c, located over the range (3 ; 4),  has to verify the identity

6c - 9 = [(c-1)(c-2)]^2,

for Lagrange's theorem to be valid.