Problem 13 - Essential Calculus

Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0\).$$ f(x)=x^{2 / 3}-1,[-8,8] $$

Step-by-Step Solution

Answer

Rolle's theorem can be applied to the function \(f(x)=x^{2/3}-1\) on the interval [-8,8], and the value of c in the interval (-8, 8) for which \(f'(c) = 0\) is c = 0.

Step 1: Check Continuity and Differentiability

First, analyze whether the function \(f(x) = x^{2/3} - 1\) is continuous and differentiable. Yet, x^{2/3} is continuous and differentiable for all real numbers. Therefore, the given function is continuous on [-8,8] and differentiable on (-8,8).

Step 2: Check Equal Function Values at a and b

The function \(f(x) = x^{2/3} - 1\) is continuous and differentiable. The theorem further requires that f(a) = f(b). In this case, a = -8 and b = 8. If you substitute -8 for x in f(x), and also substitute 8 for x in f(x), you get -1 in both cases. This means the function meets the third condition of Rolle's Theorem.

Step 3: Find the Derivative

Next, find the derivative of the given function. In this case, \(f'(x) = \frac{2}{3} * x^{-1 /3}\). Solve \(f'(x) = 0 \). This gives x = 0. Therefore, c = 0.

Key Concepts

Understanding Continuity in Calculus

Continuity is a fundamental concept in calculus that deals with the behavior of functions. For a function to be continuous at a point, it must meet three conditions: the function must be defined at the point, it must have a limit as it approaches the point, and the limit must equal the function's value at that point.

For a real-valued function like the one we have, which is the function \(f(x) = x^{2/3} - 1\), the question of continuity is about whether the graph of the function has any breaks, jumps, or holes on the interval \([-8, 8]\). If the graph can be drawn without lifting the pencil off the paper, the function is likely continuous.

As noted in our exercise case, \(x^{2/3}\) is a root function that yields real numbers for all real inputs which means there are no breaks in its graph. Subtraction of 1 does not affect this continuity. Thus, the function satisfies the first condition for applying Rolle's Theorem on the given interval.

Differentiability and Its Role

Differentiability refers to a function's ability to have a derivative at each point in its domain. A function is differentiable at a point if it has a unique tangent line at that point. In more simple terms, if you can find the slope of the function at any point without ambiguity, the function is differentiable there.

However, for a function to be differentiable on an interval, as required by Rolle's Theorem, it must be differentiable at every point within the interval (but not necessarily at the endpoints).

For the function \(f(x) = x^{2/3} - 1\), differentiability required a closer look because root functions can sometimes have points, often at zeros, where a tangent is vertical or the function isn't smooth. Despite this, the function \(x^{2/3}\) avoids these issues for all real numbers, making it differentiable on the open interval \((-8, 8)\), which meets the second criterion for Rolle's Theorem.

Finding Derivatives to Apply Rolle's Theorem

Finding derivatives is a critical process in calculus, especially when applying Rolle's Theorem. The derivative of a function at any point represents the rate at which the function's value is changing at that point—it's the slope of the tangent line to the function's graph.

To find whether a function satisfies Rolle's Theorem, we need to determine if there's a point in the interval where the derivative is zero, which would mean the function has a horizontal tangent line there (indicating a possible local maximum or minimum).

In the given exercise, the derivative \(f'(x) = \frac{2}{3} x^{-1/3}\) is found using the power rule for derivatives. Equating the derivative to zero, we solve for \(x\) to find potential values where the function's slope is zero. The solution leads us to \(x = 0\), which within the interval \((-8, 8)\), confirms the point where the function's rate of change is zero, thus satisfying the conditions of Rolle's Theorem.