Using Rolle’s Theorem, find points on the curve y = 16 – x2, x ∈

1.	Using Rolle’s Theorem, find points on the curve y = 16 – x2, x ∈ [-1, 1], where tangent is parallel to x – axis.
Answer» Given as the function is y = 16 – x², x ϵ [– 1, 1] As we know that polynomial function is continuous and differentiable over R. Let's check the values of ‘y’ at extremes ⇒ y (– 1) = 16 – (– 1)² ⇒ y (– 1) = 16 – 1 ⇒ y (– 1) = 15 ⇒ y (1) = 16 – (1)² ⇒ y (1) = 16 – 1 ⇒ y (1) = 15 y(– 1) = y(1). Therefore, there exists a c ϵ (– 1, 1) such that f’(c) = 0. As 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 us find the derivative of curve y y' = d(16 - x²)/dx ⇒ y’ = – 2x y’(c) = 0 ⇒ – 2c = 0 ⇒ c = 0 ϵ (– 1, 1) Value of y at x = 1 is ⇒ y = 16 – 0² ⇒ y = 16 Hence, 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).

Using Rolle’s Theorem, find points on the curve y = 16 – x2, x ∈ [-1, 1], where tangent is parallel to x – axis.

Answer»

Given as the function is y = 16 – x², x ϵ [– 1, 1]

As we know that polynomial function is continuous and differentiable over R.

Let's check the values of ‘y’ at extremes

⇒ y (– 1) = 16 – (– 1)²

⇒ y (– 1) = 16 – 1

⇒ y (– 1) = 15

⇒ y (1) = 16 – (1)²

⇒ y (1) = 16 – 1

⇒ y (1) = 15

y(– 1) = y(1). Therefore, there exists a c ϵ (– 1, 1) such that f’(c) = 0.

As 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 us find the derivative of curve y

y' = d(16 - x²)/dx

⇒ y’ = – 2x

y’(c) = 0

⇒ – 2c = 0

⇒ c = 0 ϵ (– 1, 1)

Value of y at x = 1 is

⇒ y = 16 – 0²

⇒ y = 16

Hence, 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).