Problem 16 - Calculus Early Transcendentals

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

There is no \( c \) because \( f'(x) \) is not defined at \( x=0 \). This doesn't contradict the MVT as \( f \) is not differentiable on \((a, b)\).

Step 1: Compute f(b) and f(a)

Calculate the values of the function at the endpoints of the interval. For \( b = 8 \): \[ f(8) = 8^{2/3} = 4 \] For \( a = -1 \): \[ f(-1) = (-1)^{2/3} = 1 \] Thus, \( f(8) = 4 \) and \( f(-1) = 1 \).

Step 2: Calculate the Average Rate of Change

Find the average rate of change of the function over the interval \( [a, b] \). The formula to calculate this is: \[ \frac{f(b) - f(a)}{b-a} = \frac{4 - 1}{8 - (-1)} = \frac{3}{9} = \frac{1}{3} \] This is the slope of the secant line connecting the points \((a, f(a))\) and \((b, f(b))\).

Step 3: Find f'(x) and analyze its existence in (a, b)

Find the derivative of \( f(x) = x^{2/3} \) and determine where it exists.Differentiate: \[ f'(x) = \frac{d}{dx}(x^{2/3}) = \frac{2}{3}x^{-1/3} = \frac{2}{3 \sqrt[3]{x}} \] Note that \( f'(x) \) does not exist at \( x = 0 \). Since \( 0 \in (-1, 8) \), there is no \( c \) in \( (a, b) \) where \( f'(c) = \frac{1}{3} \) because \( f'(x) \) is undefined at 0.

Step 4: Explain Non-contradiction with Mean-Value Theorem

The Mean-Value Theorem (MVT) states that if \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \). However, since \( f(x) = x^{2/3} \) is not differentiable at \( x = 0 \) within the interval \((-1, 8)\), the conditions for the MVT are not met. Thus, the result from (a) does not contradict the Mean-Value Theorem.

Key Concepts

Continuous Function

In mathematics, a function is described as continuous if there are no breaks, jumps, or holes in its graph. More technically, a function \( f(x) \) is said to be continuous at a point \( x = a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to the function's value at that point, \( f(a) \). For the entire function to be continuous over an interval, this condition must hold at every point within the interval, including the endpoints.

Continuity is a vital concept when applying the Mean Value Theorem (MVT), which requires the function to be continuous over the closed interval \([a, b]\). In our original exercise, the function \( f(x) = x^{2/3} \) is continuous in the given domain \((-1, 8)\). You can think of continuous functions as smooth pathways that work perfectly for applications like MVT.

To recap:

• A continuous function has no sudden changes in value over its domain.
• The value of the function at any point on its graph can be approached gradually, without any jump.

Differentiable Function

Differentiability refers to the ability of a function to have a derivative at every point in its domain. A function is differentiable over an interval if it has a derivative at every point within that interval. This means you can draw a tangent line to the curve of the function at any point in the interval.

For the Mean Value Theorem to apply, a function must be differentiable over the open interval \((a, b)\). In our exercise, the function \( f(x) = x^{2/3} \) is not differentiable at \( x = 0 \), as the derivative \( f'(x) = \frac{2}{3\sqrt[3]{x}} \) becomes undefined at \( x = 0 \).

Considerations regarding differentiability:

• A function can be continuous but not differentiable. Continuity is necessary for differentiability, but not sufficient.
• If a function is not differentiable at some point within an interval, it does not satisfy the conditions necessary for applying the Mean Value Theorem.

Average Rate of Change

The average rate of change of a function over an interval provides a way of understanding how the function behaves on average over that interval. It is computed by taking the change in the function's value at the endpoints of the interval and dividing it by the change in the endpoints themselves.

This measurement is akin to finding the slope of the line that connects the two endpoints on the graph of the function, often termed the 'secant line.' In our exercise, the average rate of change of \( f(x) = x^{2/3} \) from \( x = -1 \) to \( x = 8 \) is \( \frac{1}{3} \), calculated as follows: \( \frac{4 - 1}{8 - (-1)} = \frac{1}{3} \).

Key points to remember about the average rate of change:

• It's similar to measuring the slope between two points on the function's graph.
• Useful for estimating how the function is changing overall between two points.
• Provides a baseline to compare actual rates of change at specific points within the interval.