Problem 5 - Thomas Calculus

Question

Satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. \(f(x)=x^{2 / 3}, \quad[-1,8]\)

Step-by-Step Solution

Answer

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

Step 1: State the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 2: Identify Continuity and Differentiability Requirements

For \( f(x) = x^{2/3} \) over \([-1, 8]\), we must check if the function is continuous on \([-1, 8]\) and differentiable on \((-1, 8)\).

Step 3: Check Continuity of \( f(x) \)

\( f(x) = x^{2/3} \) is continuous for all \( x \) in the interval \([-1, 8]\), because it is a composition of the functions \( x^{2/3} \) which is continuous everywhere.

Step 4: Check Differentiability of \( f(x) \)

The derivative of \( f(x) = x^{2/3} \) is \( f'(x) = \frac{2}{3}x^{-1/3} \) or \( \frac{2}{3 \sqrt[3]{x}} \). This derivative does not exist at \( x = 0 \), so \( f(x) \) is not differentiable on the entire interval \((-1, 8)\).

Step 5: Conclusion on Satisfying MVT

Since \( f(x) \) is not differentiable at \( x = 0 \), \( f(x) \) does not satisfy the conditions of the Mean Value Theorem on the interval \([-1, 8]\).

Key Concepts

Understanding Continuity in Functions

In mathematics, the concept of continuity describes how a function behaves over a specific interval. Simply put, a function is continuous if you can draw its graph without lifting your pen from the paper. If we consider the function \( f(x) = x^{2/3} \), we need to determine its continuity on the interval \([-1, 8]\).

The function \( f(x) = x^{2/3} \) is a composition of a power function, which is generally continuous everywhere in its domain. Since there's no break or gap in the function across \([-1, 8]\), it maintains continuity over this closed interval. The importance of checking for continuity lies in many calculus theorems that require it, especially the Mean Value Theorem.

Key points about continuity:

• A continuous function has no sudden jumps or holes.
• For the Mean Value Theorem to apply, continuity must be satisfied over the closed interval \([a, b]\).
• Verification of continuity is the first step in applying numerous calculus theorems effectively.

Understanding Differentiability in Functions

Differentiability refers to the existence of a function's derivative at every point in its domain. For a function to be differentiable at a point, it must be smooth without any sharp vertices or cusps. The function \( f(x) = x^{2/3} \) might cause curiosity here since its graph peaks at \( x = 0 \).

Calculating the derivative can reveal whether \( f(x) = x^{2/3} \) is differentiable in the interval \((-1, 8)\). We find that the derivative is \( f'(x) = \frac{2}{3 \sqrt[3]{x}} \). A crucial observation is the derivative's non-existence at \( x = 0 \) because it results in division by zero.

Key points about differentiability:

• A function is differentiable if it possesses a derivative at all points.
• Non-differentiable points might be due to gaps, sharp turns, or vertical tangents.
• Differentiability is required on the open interval \((a, b)\) for the Mean Value Theorem.

Interval Analysis for Calculus Theorems

Interval analysis involves examining a function over a specific range of values to apply various calculus theorems effectively. For the Mean Value Theorem, it emphasizes both continuity and differentiability over the respective closed \([a, b]\) and open \((a, b)\) intervals.

When we evaluate \( f(x) = x^{2/3} \) over the interval \((-1, 8)\), we must ensure that its properties align with the conditions required by the theorem. While continuity is satisfied across the closed interval, differentiability fails at \( x = 0 \), failing over the entire open interval \((-1, 8)\).

Key aspects of interval analysis:

• It ensures functions meet the assumptions of pivotal theorems.
• Easily identifies whether conditions like continuity and differentiability are met on given intervals.
• Awareness of potential issues, like non-differentiability points, is crucial for proper theorem application.