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 = 16 – x², xϵ[ – 1,1]

We know that polynomial function is continuous and differentiable over R.

Let’s check the values of ‘y’ at extremums

⇒ y( – 1) = 16 – ( – 1)²

⇒ y( – 1) = 16 – 1

⇒ y( – 1) = 15

⇒ y(1) = 16 – (1)²

⇒ y(1) = 16 – 1

⇒ y(1) = 15

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

We know that for a curve g, the value of the slope of the tangent at a point r is given by g’(r).

Let’s find the derivative of curve y

⇒

⇒ y’ = – 2x

We have y’(c) = 0

⇒ – 2c = 0

⇒ c = 0ϵ( – 1,1)

Value of y at x = 1 is

⇒ y = 16 – 0²

⇒ y = 16

∴ The point at which the curve y has a tangent parallel to x – axis (since the slope of x – axis is 0) is (0,16).