Problem 9 - University Calculus: Early Transcendentals

Question

Which of the functions 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 MVT as it is not differentiable at \( x = 0 \).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \( f \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a point \( c \) in the interval \( (a, b) \) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]The function must be both continuous on the closed interval and differentiable on the open interval for the theorem to apply.

Step 2: Test for Continuity on \([-1, 8]\)

The function \( f(x) = x^{2/3} \) is continuous everywhere on \(-\infty, \infty\) because it is defined for all \( x \) in the real numbers. Therefore, \( f(x) \) is certainly continuous on the closed interval \([-1, 8]\).

Step 3: Test for Differentiability on \((-1, 8)\)

A function is differentiable at a point if its derivative is defined at that point. Calculate the derivative of \( f(x) = x^{2/3} \):\[ f'(x) = \frac{2}{3} x^{-1/3} \]For this derivative to be defined, \( x \) must be positive because \( x^{-1/3} \) becomes problematic (undefined) at \( x = 0 \) (division by zero). Thus, \( f'(x) \) is not defined on the interval \((-1, 8)\) at \( x = 0 \). Therefore, \( f(x) \) is not differentiable on \((-1, 8)\).

Step 4: Conclusion

Since \( f(x) = x^{2/3} \) is not differentiable at \( x = 0 \) within the interval \((-1, 8)\), it does not satisfy the differentiability requirement of the Mean Value Theorem. Therefore, the function does not satisfy the hypotheses of the Mean Value Theorem on the given interval.

Key Concepts

Continuity

Continuity is a key requirement for applying the Mean Value Theorem. When we say a function is continuous over a closed interval \([a, b]\), it means that you can draw the graph of the function from point \((a, f(a))\) to point \((b, f(b))\) without lifting your pencil. Essentially, a continuous function has no breaks, jumps, or holes in this interval.
To determine if a function is continuous over a given interval, check the following:

• The function is defined at every point within the interval.
• The limit of the function as it approaches any point within the interval from either side must exist.
• The function's value and its limit at each point within the interval must be equal.

For example, the function \(f(x) = x^{2/3}\) is continuous over the interval \([-1, 8]\) because it is defined for all real numbers, and the smooth progression of its graph reflects continuity over this interval.

Differentiability

Differentiability is another crucial requirement for the Mean Value Theorem. A function is differentiable at a point if it has a well-defined tangent at that point, meaning you can find its derivative there. Differentiability implies a smooth graph without any sharp corners or cusps.
When checking differentiability over an open interval \( (a, b) \), consider:

• The derivative must exist at each point within the interval.
• The function must not have any vertical tangents within the interval.
• There should be no abrupt changes in the direction of the function.

For \(f(x) = x^{2/3}\), the derivative is \(f'(x) = \frac{2}{3}x^{-1/3}\). Here, differentiability fails at \(x = 0\) because \(x^{-1/3}\) leads to division by zero, which is undefined. Therefore, despite being continuous, the function is not differentiable on \(( -1, 8 )\). This crucial gap in differentiability affects satisfying the Mean Value Theorem requirements.

Function Analysis

Function analysis helps us understand the behavior and properties of functions. For the application of the Mean Value Theorem, identifying whether a function meets the continuity and differentiability requirements is essential. Function analysis dives deeper into:

• Examining any points of discontinuity or non-differentiability.
• Determining where the function's behavior can violate theorem conditions.
• Assessing global and local behavior to predict function behavior in specific intervals.

In analyzing \(f(x) = x^{2/3}\), while function analysis confirms continuity on \([-1, 8]\), it reveals a critical lapse at \(x = 0\), rendering the function non-differentiable there. This insight is vital since the Mean Value Theorem demands uninterrupted differentiability. Understanding these nuances in function analysis can illuminate why a function like \(x^{2/3}\) might fail to fulfill the theorem's hypotheses, even when appearing smooth and intact at first glance.