Q. 724.0( 3 Votes )

Using Rolle’s theorem, find the point on the curve y = x(x – 4), where the tangent is parallel to x-axis.

Answer :

Given: y = x(x – 4)

y = (x2 – 4x)


Now, we have to show that f(x) verify the Rolle’s Theorem


First of all, Conditions of Rolle’s theorem are:


a) f(x) is continuous at (a,b)


b) f(x) is derivable at (a,b)


c) f(a) = f(b)


If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0


Condition 1:


On expanding y = x(x – 4), we get y = (x2 – 4x)


Since, x2 – 4x is a polynomial and we know that, every polynomial function is continuous for all x R


y = (x2 – 4x) is continuous at x [0,4]


Hence, condition 1 is satisfied.


Condition 2:


y = (x2 – 4x)


y’ = 2x – 4


x2 - 4x is differentiable at [0,4]


Hence, condition 2 is satisfied.


Condition 3:


y = x2 – 4x


when x = 0


y = 0


when x = 4


y = (4)2 – 4(4) = 16 – 16 = 0


Hence, condition 3 is also satisfied.


Now, there is atleast one value of c (0,4)


Given tangent to the curve is parallel to the x – axis


This means, Slope of tangent = Slope of x – axis



2x - 4 = 0


2x = 4


x = 2 (0, 4)


Put x = 2 in y = x2 – 4x , we have


y = (2)2 – 4(2) = 4 – 8 = -4


Hence, the tangent to the curve is parallel to the x –axis at


(2, -4)


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