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

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.