f(x) = (2x-1)/(x+5) the inteval [-1,1]

Aaccorging Lagrange theorem:

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

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

f(1) f(-1) = 2*f'(c)

Let us calculate:

f(1) = (2-1)/(1+5) = 1/6

f(-1) = (-2-1)/(-1+5) = -3/4

Now let us differentiate f(x)

f(x) = (2x-1)/(x+5) = u/v

u= 2x-1 ==> u' = 2

v= x+5 ==> v' = 1

==> f'(x) = (u'v- uv')/v^2

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

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

= 11/(x+5)^2

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

Now let us substitute in the equation:

f(1) - f(-1) = 2*f'(c0

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

1/6 + 3/4 = 22/(c+5)^2

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

11/12 = 22/(c^2 + 10c + 25)

cross multiply:

==> 11(c^2 + 10 c + 25) = 12*22

==> 11c^2 + 110c + 275 = 264

==> 11c^2 + 110c + 11 = 0

Divide by 11:

==> c^2 + 10c + 1 = 0

==> **c1= [-10 + sqrt(100-4)]/2 = -5 + 2sqrt6**

**==> c2= -5 - 2sqrt6**

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.**

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

To apply Legrange's theorem.

We know that in the interval (a, b) if f(x) is cntinous and derivable, then

there exist a point c in (a, b) such that f'(c) = (f(b)-f(a))/(b-a).

Here f(-1) = (2*(-1) -1)/(-1+5) = -3/4 .

f(1) = (2*1-1)/(1+5) = 1/6

(f(1)-f(-1) )/(1- -1) = (1/6)- - 3/4)/(1- -1) = (4+18)/(24*2) = 22/48 = 11/24.

Now let us finf f'(x) = (2x-1)/(x+5)' = {(2x-1)'(x+5) - (2x-1)(x+5)'}/(x+5)^2.

= {2(x+5) -(2x-1)(1)}/(x+5)^2

= {2x+10-2x+1}/(x+5)^2

=11/(x+5)^2 .

f'(x) = 11/(x+5)^2 if this is equal to 11/24, then

11/(x+5)^2 = 11/24. Or

(x+5)^2 = 24

x+5 = sqrt 24

x = 5+sqrt24 or x 5-sqrt24 = 0.1010... is a value in (-1, 1) which verifies the Legranges theorem.