Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ g(x)=x^{5 / 3} ;[0,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply since the function is not differentiable at \(x=0\).

Step 1: Verify Continuity

The first condition of the Mean Value Theorem (MVT) is that the function must be continuous on the closed interval \([0, 1]\). The function \(g(x) = x^{5/3}\) is a power function, which is continuous for all real numbers including the interval \([0, 1]\).

Step 2: Verify Differentiability

The second condition of the MVT is that the function must be differentiable on the open interval \((0, 1)\). We calculate the derivative of \(g(x)\): \[ g'(x) = \frac{5}{3}x^{2/3} \] This derivative is undefined at \(x = 0\), which means that the function is not differentiable at \(x = 0\), failing the differentiability requirement on the entire open interval since \(0\) is included.

Step 3: Apply Mean Value Theorem Criteria

Since \(g(x) = x^{5/3}\) is not differentiable on the entire open interval \((0, 1)\) due to the point \(x=0\), the Mean Value Theorem does not apply to this function on the interval \([0, 1]\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus. A function is continuous on a closed interval when there are no breaks, jumps, or holes in its graph within that interval. Essentially, you can draw the graph of the function within this interval without lifting your pencil from the paper. For power functions like \(g(x) = x^{5/3}\), continuity often comes naturally, because these functions are defined for all real numbers. In our example, \(g(x)\) is continuous on the interval \([0, 1]\) because power functions of the form \(x^n\) are continuous everywhere. This aspect is important for the Mean Value Theorem (MVT) which requires the function to be continuous on a closed interval to apply. Continuity ensures that all values of \(x\) in the interval can smoothly connect through \(g(x)\).

Differentiability

Differentiability is about whether a function has a derivative at every point within an interval. A function is differentiable on an interval if it has a defined slope at each point within that interval. This slope is described by the derivative. The Mean Value Theorem requires a function to be differentiable on an open interval to ensure the continuous change of slope. Calculating the derivative of our function \(g(x) = x^{5/3}\), we have \[ g'(x) = \frac{5}{3}x^{2/3} \]. However, this derivative is not defined at \(x = 0\), because \(x^{2/3}\) doesn't allow \(x = 0\) as it creates an undefined expression in the context of meaningful calculus. As a result, although the function is mostly differentiable on \((0, 1)\), the presence of \(x = 0\) causes a problem, preventing the use of the MVT over this interval.

Power Functions

Power functions are a type of function where the dependent variable is equal to a variable base raised to some power, like \(f(x) = x^n\). These functions are characterized by their simplicity and wide applicability in calculus. When considering continuity and differentiability:

• Power functions are generally continuous everywhere because they have no abrupt changes in their middle or endpoints.
• However, differentiability can be more restrictive, especially when the exponent is a fraction like in \(x^{5/3}\). At integral points where the function's derivative doesn't exist, differential calculus tools cannot be applied, complicating the use of the MVT.

Understanding these characteristics helps anticipate issues in analysis and problem-solving in calculus exercises.

Intervals in Calculus

Intervals play a crucial role in calculus as they define the portion of a function we are analyzing. They are of two types: closed and open intervals. A closed interval, like \([0, 1]\), includes its endpoints, meaning the values at \(0\) and \(1\) are considered in the analysis. In contrast, an open interval, such as \((0, 1)\), does not include its endpoints because it looks purely at the interval's internal behavior.Both types of intervals are essential in applying the Mean Value Theorem. For MVT to apply, the function in question must be continuous on the closed interval and differentiable on the open interval. In our example, while \(g(x) = x^{5/3}\) is continuous over \([0, 1]\), it is not differentiable at \(x=0\) over \((0, 1)\) leading to issues in meeting the MVT criteria. Proper understanding of these intervals assists in predicting how the functions behave across different scenarios.