First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.

Given function is:

⇒ y = 12(x + 1)(x – 2) on [ – 1,2]

We know that polynomials are continuous and differentiable over R.

Let’s check the values of y at the extremums

⇒ y( – 1) = 12( – 1 + 1)( – 1 – 2)

⇒ y( – 1) = 12(0)( – 3)

⇒ y( – 1) = 0

⇒ y(2) = 12(2 + 1)(2 – 2)
⇒ y(2) = 12(3)(0)

⇒ y(2) = 0

We got y( – 1) = y(2). So, there exists a c such that f’(c) = 0.

For a curve g to have a tangent parallel to the x – axis at point r, the criteria to be satisfied is g’(r) = 0.

⇒ y’(x) = 0

⇒

⇒

⇒ ((x + 1)×1) + ((x – 2)×1) = 0

⇒ x + 1 + x – 2 = 0

⇒ 2x – 1 = 0

⇒ 2x = 1

⇒

The value of y is

⇒

⇒

⇒ y = – 27

The point at which the curve has tangent parallel to x – axis is .