Question

Using Lagrange's Mean Value theorem find a point on the curve y=(x-3)^2 where tangent is parallel to chord joining (3,0) and (4,1).

Answer 1

Lagrange''s  theorem states that for a function  f (x) = y,  defined over [a, b] which is continuous in the above range and the derivative f '(x) is defined over (a , b) :

 there exists a point X, at which : the derivative is such that:

     f '(X) = [ f (b) - f(a) ] / (b - a)  = slope of the function at x = X

   given   f (x) =  (x - 3)²
             f '(x)  = 2 (x - 3)    ---- (1)

the interval is  x = [ 3, 4 ]
     b = 4  and  a = 3
     f(a) = 0    and    f(b) = 1

   f '(X) =  [ 1 - 0 ]  / [ 4 - 3]   = 1

   So  using equation (1),   2 (x - 3) = 1
             =>  x = 7/2
  The point on the curve y = (x - 3)²  needed is (7/2, 1/4)