Problem 181 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b]\). Justify your answer. \(y=5+|x|\) over [-1,1]

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because the function is not differentiable at \( x = 0 \).

Step 1: Check Continuity over the Interval

The function \( y = 5 + |x| \) is a combination of constant addition and absolute value. The absolute value function is continuous everywhere, and adding a constant doesn't affect continuity. Therefore, \( y = 5 + |x| \) is continuous over the interval \([-1, 1]\).

Step 2: Check Differentiability over the Interval

The function \( y = 5 + |x| \) needs to be differentiable over \((-1, 1)\) to satisfy the Mean Value Theorem. We find the derivative at points away from where \( x = 0 \). On \((0, 1)\), \( y = 5 + x \) so the derivative is 1. On \((-1, 0)\), \( y = 5 - x \) so the derivative is -1. At \( x = 0 \), the derivative does not exist because the left-hand limit of the derivative (-1) does not equal the right-hand limit (1).

Step 3: Conclusion on Applicability of Mean Value Theorem

For the Mean Value Theorem to apply, the function must be continuous on \([a, b]\) and differentiable on \((a, b)\). While \( y = 5 + |x| \) is continuous on \([-1, 1]\), it is not differentiable at \( x = 0 \). Therefore, the Mean Value Theorem does not apply.

Key Concepts

Continuity

Continuity is an essential idea in calculus, specially when utilizing the Mean Value Theorem. A function is continuous over an interval if you can draw it without lifting your pencil off the paper through that interval. In simpler terms, there are no gaps, jumps, or holes in the graph of the function. For the function \( y = 5 + |x| \) over the interval \([-1, 1]\), continuity is ensured because:

• The absolute value function \(|x|\) is inherently continuous for all real numbers.
• Addition of a constant, like 5, does not disrupt continuity.

Hence, we can confirm that \( y = 5 + |x| \) is continuous across our selected interval, making the first criterion of the Mean Value Theorem satisfied.

Differentiability

Differentiability is a step further than continuity. A function is differentiable in an interval if its derivative exists at each point in that interval. In the context of the Mean Value Theorem, differentiability is required in the open interval \((a, b)\). For our function \( y = 5 + |x| \) over \((-1, 1)\):

• On \((0, 1)\), the function simplifies to \( y = 5 + x \), with a derivative of 1.
• On \((-1, 0)\), it simplifies to \( y = 5 - x \), with a derivative of -1.
• At \( x = 0 \), both left-hand and right-hand derivatives differ, therefore, the derivative does not exist at this point.

Since the function is not differentiable at \( x = 0 \), it fails to meet the differentiability requirement of the Mean Value Theorem, rendering the theorem inapplicable over \([-1, 1]\).

Piecewise Function

Piecewise functions are defined by multiple sub-functions, each covering a part of the function's domain. These functions can lead to interesting behaviors, particularly at boundaries (or breakpoints) between different pieces. The function \( y = 5 + |x| \) can be interpreted as a piecewise function:

• For \( x \geq 0 \), \( y = 5 + x \).
• For \( x < 0 \), \( y = 5 - x \).

This definition presents a corner at \( x = 0 \) on the graph, making the derivative undefined at this point. While the function remains continuous, it does not guarantee differentiability at the junction point, contributing to the failure of the Mean Value Theorem's applicability.

Calculus Problem-Solving

Problem-solving in calculus involves systematically applying important theorems and methods. When determining if the Mean Value Theorem applies, we begin by ensuring two key conditions:

• Continuity over the closed interval \([a, b]\).
• Differentiability over the open interval \((a, b)\).

For \( y = 5 + |x| \) over \([-1, 1]\), the continuity condition is met, but differentiability fails at \( x = 0 \). Consequently, the theorem does not apply. Successfully solving such problems involves recognizing these conditions, calculating derivatives where needed, and interpreting piecewise function behavior. These skills ensure precise and accurate conclusions, fostering a deeper understanding of calculus principles.