Apply Lagrange theorem to f = (2x-1)/(x+5) in the interval [-1,1]

We can use Lagrange theorem if and only if the function is
continuous. We'll analyze the function f(x) = (2x-1)/(x+5) which is a function that has
the numerator and denominator, elementary functions. Because of the fact that the
elementary functions are continuously, f(x) is a continuous function,
also.

Because of the fact that f(x) is a continuous
function, over the interval [-1 , 1], that means that it could be differentiated over
the interval (-1,1).

According to Lagrange teorem, there is
a point c, that belongs to the interval (-1,1), so
that:

f(1) - f(-1) =
f'(c)(1+1)

We'll calculate
f(1):

f(1) =
(2*1-1)/(1+5)

f(1) = 1/6

We'll
calculate f(-1):

f(-1) =
(2*(-1)-1)/(-1+5)

f(-1) =
-3/4

We'll differentiate f(x). Since f(x) is a ratio, we'll
differentiate using the quotient rule:

(u/v)' = (u'*v -
u*v')/v^2

f'(x) = [(2x-1)'*(x+5) -
(2x-1)*(x+5)']/(x+5)^2

f'(x) = [2(x+5) - 2x +
1]/(x+5)^2

f'(x) =
(2x+10-2x+1)/(x+5)^2

We'll eliminate like
terms:

f'(x) = 11/(x+5)^2

So,
f'(c) = 11/(c+5)^2

We'll re-write Lagrange
theorem:

f(1) - f(-1) =
f'(c)(1+1)

1/6 + 3/4 =
2*11/(c+5)^2

We'll calculate LCD to the left
side:

1/6 + 3/4 =
(2+3*3)/12

1/6 + 3/4 =
11/12

11/12 = 22/(c+5)^2

We'll
divide by 11:

1/12 =
2/(c+5)^2

We'll cross
multiply:

24 = (c+5)^2

We'll
expand the square:

c^2 + 10c + 25 - 24 =
0

c^2 + 10c + 1 = 0

We'll
apply the quadratic formula:

c1 =
[-10+sqrt(100-4)]/2

c1 =
(-10+4sqrt6)/2

c1 =
2(-5+2sqrt6)/2

c1 =
-5+2sqrt6

c1 = -0.12 approx

c2
= -5-2sqrt6

c2 = -9.88
approx

Since c has to belong to the interval
(-1,1), the only admissible value is c1 =
-5+2sqrt6.