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)