According to the Mean Value Theorem (http://en.wikipedia.org/wiki/Mean_value_theorem), if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that      =  .

  F(x) =3+2x+5.

(x) = 3(2x) + 2= 6x + 2.

  Let c be a point in ( a,b ) such that.

  6c + 2 =0è 6c = -2 => c = = , c is to lie in (a,b).

  At x=a è f(a) = 3 + 2a + 5.

  At x = b è f(b) = 3 + 2b + 5 .

  Then    =   =  = =

  Now, we know from the line above (highlighted in red) that (c) = 6c + 2.

  So 6c + 2 = è 6c = 3(b+a)+2-2.

  6c =3(b+a)+2-2

  6c = 3(b + a).

  c = = .

  So this shows that for any interval on this curve, there is a value of c (equal to ) that satisfies the theorem.

  At this value of c,   =