Q. 85.0( 2 Votes )

Find points on the curve y = x3 – 3x, where the tangent to the curve is parallel to the chord joining (1, – 2) and (2, 2).

Answer :

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that


f(b)−f(a)=f′(c)(b−a)



This theorem is also known as First Mean Value Theorem.


Let f(x) = x3 – 3x on [1, 2]


This interval [a, b] is obtained by x – coordinates of the points of the chord.


Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.


Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.





f(x) = x3 – 3x


Differentiating with respect to x:


f’(x) = 3x2 – 3


For f’(c), put the value of x=c in f’(x):


f’(c)= 3c2 – 3


For f(2), put the value of x=2 in f(x):


f(2)= (2)3 – 3(2)


= 8 – 6


= 2


For f(1), put the value of x=1 in f(x):


f(1) = (1)3 – 3(1)


= 1 – 3


= – 2


f’(c) = f(2) – f(1)


3c2 – 3 = 2 – ( – 2)


3c2 – 3 = 2 + 2


3c2 = 4 + 3





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 (1, – 2) and (2, 2).


Now, Put this value of x in f(x) to obtain y:


y = x3 – 3x








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