Given as f(x) = (x – 4)2 on [4, 5]
In this interval [a, b] is obtained by x – coordinates of the points of the chord.
The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Since, f(x) is a polynomial function. Therefore it is continuous in [4, 5] and differentiable in (4, 5). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied.
So, there exist a point c ∈ (4, 5) such that:
Differentiate with respect to x
f'(x) = 2(x - 4)(d(x - 4)/dx)
⇒ f’(x) = 2 (x – 4) (1)
⇒ f’(x) = 2 (x – 4)
For the f’(c), put the value of x=c in f’(x):
f’(c) = 2 (c – 4)
For the f(5), put the value of x=5 in f(x):
f (5) = (5 – 4)2
= (1)2
= 1
For the f(4), put the value of x=4 in f(x):
f (4) = (4 – 4)2
= (0)2
= 0
f’(c) = f(5) – f(4)
⇒ 2(c – 4) = 1 – 0
⇒ 2c – 8 = 1
⇒ 2c = 1 + 8
c = 9/2 = 4.5 ∈ (4,5)
As we know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (4, 0) and (5, 1).
Putting this value of x in f(x) to obtain y:
y = (x – 4)2
Thus, the required points is (9/2, 1/4)