Problem 154 - Calculus Volume 1

Question

Determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer. $$ y=\sqrt{4-x^{2}} $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies on the interval \([-2, 2]\).

Step 1: Identify the Function and Interval

We start with the function given: \[ y = \sqrt{4 - x^2} \]This function is defined for all values of \( x \) where the expression under the square root is non-negative, i.e., \( 4 - x^2 \geq 0 \). The inequality simplifies to \(-2 \leq x \leq 2\), dictating that this function is defined on the interval \([-2, 2]\).

Step 2: Check Continuity

The function \( y = \sqrt{4 - x^2} \) is continuous on the open interval \((-2, 2)\) since it's composed of continuous operations: subtraction and square root. At the endpoints \( x = -2 \) and \( x = 2 \), \( y \) is also defined, making it continuous on the closed interval \([-2, 2]\).

Step 3: Check Differentiability

To ensure the Mean Value Theorem applies, the function must also be differentiable on the open interval \((-2, 2)\).We compute the derivative using the chain rule:\[ y' = \frac{d}{dx} \sqrt{4 - x^2} = \frac{-x}{\sqrt{4 - x^2}} \]The derivative is well-defined and continuous on \((-2, 2)\), as the square root in the denominator is non-zero for \(-2 < x < 2\). Therefore, \( y \) is differentiable on \((-2, 2)\).

Step 4: Confirm Conditions for Mean Value Theorem

Since the function is continuous on \([-2, 2]\) and differentiable on \((-2, 2)\), all conditions for the Mean Value Theorem are satisfied. Therefore, the Mean Value Theorem applies over the interval \([-2, 2]\).

Key Concepts

Continuity

In mathematics, continuity is a crucial concept that plays a significant role in ensuring smooth and unbroken behavior of functions. To say a function is continuous on an interval means there are no abrupt changes or jumps at any point within that interval. For the function given in the exercise, \( y = \sqrt{4 - x^2} \), we can determine where it is continuous by looking at the operations that define it.

• **Subtraction**: The expression \( 4 - x^2 \) is a polynomial, and polynomials are continuous everywhere. This means we aren't introducing any discontinuous behavior here.
• **Square root**: The domain of the square root function is limited to non-negative numbers, so \( 4 - x^2 \geq 0 \). This results in the interval \(-2 \leq x \leq 2\).

On the interval \([-2, 2]\), the function is continuous, as there are no points within these bounds where the function would exhibit a jump or undefined behavior.

Differentiability

Differentiability is another fundamental characteristic of functions, closely linked to the concept of having a derivative. For a function to be differentiable at a point, it essentially needs to be smooth at that location, without any sharp corners or cusps. For this specific exercise, we confirm differentiability for \( y = \sqrt{4-x^2} \) by examining the derivative we obtained: \( y' = \frac{-x}{\sqrt{4 - x^2}} \). Differentiability can be checked by ensuring that:

• The derivative exists at all points within the interval \((-2, 2)\).
• The expression under the square root, \( 4 - x^2 \), is never zero in this open interval, making it impossible for the derivative to become undefined.

This derivative doesn't indicate any sharp turns or cusps within the interval \((-2, 2)\). Hence, the function is differentiable over this interval, which satisfies one of the necessary conditions for the Mean Value Theorem to apply.

Intervals

Intervals are segments of the real line defined by starting and ending points, and they can be either open or closed. Understanding the nature of intervals is vital when deciding the applicability of calculus principles, such as the Mean Value Theorem. In the context of the function \( y = \sqrt{4 - x^2} \), it's essential to correctly identify the interval over which the Mean Value Theorem can be applied. Here, we deal with these intervals:

• **Closed Interval**: \([-2, 2]\) includes its endpoints and covers the domain where the function is defined and continuous.
• **Open Interval**: \((-2, 2)\) does not include the endpoints and is crucial when discussing differentiability. The function's derivative is valid throughout this interval.

Therefore, the Mean Value Theorem applies on the closed interval \([-2, 2]\) as the function is continuous here and on its open sub-interval \((-2, 2)\) since it's differentiable there.