Problem 16 - Calculus - AP Edition

Question

Let \(f(x)=x^{2 / 3}, a=-1,\) and \(b=8\) (a) Show that there is no point \(c\) in \((a, b)\) such that $$ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} $$ (b) Explain why the result in part (a) does not contradict the Mean-Value Theorem.

Step-by-Step Solution

Answer

No point \( c \) exists due to lack of differentiability at \( x = 0 \).

Step 1: Compute the derivative of f(x)

First, we need to find the derivative of the function \( f(x) = x^{2/3} \).Using the power rule: \( \frac{d}{dx} x^n = nx^{n-1} \), we have:\[ f'(x) = \frac{d}{dx} (x^{2/3}) = \frac{2}{3}x^{-1/3} \]

Step 2: Determine the average rate of change from a to b

The average rate of change of \( f(x) \) over the interval \( [a, b] \) is given by:\[ \frac{f(b) - f(a)}{b-a} \]Plug in \( a = -1 \) and \( b = 8 \):\[ f(8) = 8^{2/3} = 4 \]\[ f(-1) = (-1)^{2/3} = 1 \]Thus, the average rate is:\[ \frac{4 - 1}{8 - (-1)} = \frac{3}{9} = \frac{1}{3} \]

Step 3: Set derivative equal to the average rate and solve

For a point \( c \) to exist in \( (a, b) \) such that \( f'(c) = \frac{1}{3} \), we need:\[ \frac{2}{3} c^{-1/3} = \frac{1}{3} \]Solving this equation:Multiply both sides by 3:\[ 2c^{-1/3} = 1 \]Multiply both sides by \( c^{1/3} \):\[ 2 = c^{1/3} \]Cube both sides:\[ c = 8 \]However, \( c = 8 \) is not in the open interval \( (-1, 8) \), showing no such \( c \) exists.

Step 4: Explain the non-contradiction with Mean-Value Theorem

The Mean-Value Theorem (MVT) requires that the function is continuous on \([a, b]\) and differentiable on \((a, b)\). The function \( f(x) = x^{2/3} \) is continuous on \([-1, 8]\), but \( f'(x) = \frac{2}{3}x^{-1/3} \) is not defined at \( x = 0 \) due to division by zero.Since \( f(x) \) is not differentiable at \( x = 0 \), the requirements of the MVT are not fully satisfied, so there is no contradiction with the MVT.

Key Concepts

Continuity

In analyzing the Mean Value Theorem, understanding continuity is essential. A function is said to be continuous on an interval if there are no breaks, jumps, or holes throughout the interval. For a formal approach, a function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. In the context of the exercise, the function given is \( f(x) = x^{2/3} \). This function is continuous on the interval \([-1, 8]\) because you can trace the curve without lifting your pencil. Every point on the function between \(-1\) and \(8\) has a corresponding output value, and there are no sudden jumps. This continuity satisfies one of the two requirements for applying the Mean Value Theorem.

Differentiability

Differentiability refers to whether a function has a defined derivative at each point in an interval. For the Mean Value Theorem to apply, the function must be differentiable on the open interval \((a, b)\). In the given exercise, although the function \( f(x) = x^{2/3} \) is continuous on \([-1, 8]\), it is not differentiable at every point within that interval. Specifically, the derivative \( f'(x) = \frac{2}{3}x^{-1/3} \) is not defined at \( x = 0 \) due to division by zero. This gap in differentiability means that one critical condition for the Mean Value Theorem is not met, which is why the theorem does not necessarily apply to the entire interval. In this particular exercise, this discontinuity at \( x = 0 \) results in the absence of a point \( c \) that fulfills the Mean Value Theorem's criteria.

Power Rule

The power rule is a fundamental concept in calculus, used to find the derivative of a function when it's expressed as a power of \( x \). If you have a term \( x^n \), then its derivative is given by taking the exponent \( n \), multiplying the function \( x^n \) by \( n \), and subtracting \( 1 \) from the exponent. Formally, this is expressed as \( \frac{d}{dx} x^n = nx^{n-1} \). In the exercise, to find the derivative of \( f(x) = x^{2/3} \), the power rule is applied, resulting in \( f'(x) = \frac{2}{3}x^{-1/3} \). Understanding how to apply the power rule helps students easily differentiate polynomial functions, crucial in solving calculus problems similar to this exercise. It is worth noting, however, that while the power rule simplifies finding derivatives, it's also important to confirm differentiability over the required intervals when applying the Mean Value Theorem.