Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

, 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 applies; \(c \approx 0.464\) in \((0,1)\).

Step 1: Verify Applicability Conditions

The Mean Value Theorem (MVT) requires that the function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). For \(g(x) = x^{5/3}\) on \([0,1]\), we first check if it is continuous. Because \(x^{5/3}\) is a polynomial function, it is continuous everywhere, including on \([0,1]\). Next, we need to check differentiability on \((0,1)\). The derivative, \(g'(x) = \frac{5}{3}x^{2/3}\), is continuous for \(x > 0\); hence, \(g(x)\) is differentiable on \((0,1)\). MVT is applicable if these conditions are satisfied.

Step 2: Calculate Derivative

The next step is to calculate the derivative of \(g(x)\). Using the power rule, the derivative of \(g(x) = x^{5/3}\) is computed as follows:\[g'(x) = \frac{d}{dx}x^{5/3} = \frac{5}{3}x^{2/3}\]This derivative \(g'(x)\) will be used to find the value \(c\) as required by the Mean Value Theorem.

Step 3: Apply Mean Value Theorem Formula

According to MVT, there exists a \(c\) in \((0,1)\) such that:\[g'(c) = \frac{g(1) - g(0)}{1 - 0}\]The values \(g(1) = 1^{5/3} = 1\) and \(g(0) = 0^{5/3} = 0\) provide us with:\[g'(c) = \frac{1 - 0}{1} = 1\]Hence, we need to solve the equation \(\frac{5}{3}c^{2/3} = 1\) to find the required \(c\).

Step 4: Solve for c

Solving \(\frac{5}{3}c^{2/3} = 1\), we multiply both sides by \(\frac{3}{5}\) to isolate \(c^{2/3}\):\[c^{2/3} = \frac{3}{5}\]Now, raise both sides to the power of \(\frac{3}{2}\) to solve for \(c\):\[c = \left(\frac{3}{5}\right)^{3/2}\]Therefore, \(c\) is \(\left(\frac{3}{5}\right)^{3/2}\), approximately \(0.464\), which lies in the interval \((0, 1)\).

Key Concepts

Continuous Functions

A continuous function is one where small changes in the input produce small changes in the output. In other words, its graph can be drawn without any breaks or gaps. This property is crucial for the Mean Value Theorem (MVT), which requires that a function must be continuous on a closed interval \([a, b]\). For the function \(g(x) = x^{5/3}\), it is important to determine whether it is continuous over the given interval. A key fact about polynomial functions, or powers of polynomials, like \(x^{5/3}\), is that they are continuous everywhere. This means that there are no breaks in their graphs, which is evident on the interval \([0,1]\). In calculus problems involving the MVT, confirming continuity is the first step. Once we know that \(g(x)\) is continuous on \([0,1]\), we can move on to check if it is differentiable on the open interval \( (0,1) \).

Differentiability

Differentiability refers to a function's ability to have a derivative at every point in its domain. A differentiable function has a graph where there are no sharp corners or cusps, allowing us to find a tangent at every point. In the context of the Mean Value Theorem, the function must be differentiable on the open interval \( (a, b) \).For the function \(g(x) = x^{5/3}\), we compute its derivative to determine differentiability:

• Using the power rule, the derivative is \( g'(x) = \frac{5}{3}x^{2/3} \).
• This derivative \( g'(x)\) is well-defined for \( x > 0 \).

On the interval \( (0,1) \, g(x)\) is differentiable, as the derivative is continuous.Ensuring differentiability in calculus problems is key to applying the MVT, as without it, the theorem does not hold. Once confirmed, we can use the derivative to find the value \( c \) specified by the theorem.

Polynomial Functions

Polynomial functions are foundational in calculus, characterized by expressions involving powers of variables with constant coefficients. They are particularly straightforward to work with due to their smooth and continuous nature. For example, the function \(g(x) = x^{5/3}\) is technically not a polynomial because of the fractional exponent, but it behaves similarly in terms of continuity and differentiability.Polyonomial-like functions, particularly those defined by non-integer exponents, are still continuous and often differentiable on their domains. This makes them an ideal choice when working with concepts like the Mean Value Theorem. They readily satisfy the criteria of being continuous over closed intervals and differentiable over open intervals.For mathematical exercises involving polynomial functions, always remember:

• Check that the entire polynomial or polynomial-like expression is continuous and differentiable.
• Polyomials are always differentiable where their derivative does not involve division by zero.

These properties simplify the process of applying calculus theorems across various problems.

Calculus Problems

Calculus problems often require a deep understanding of continuous and differentiable functions. The Mean Value Theorem provides a powerful tool for these problems by linking the average rate of change over an interval to the instantaneous rate of change at a point within that interval.For any function defined on an interval, the procedure to solve a calculus problem using MVT involves:

• Ensuring continuity on the closed interval.
• Confirming differentiability on the open interval.
• Using the derived formula of MVT to find specific points \( c \) within the interval where the instantaneous rate of change equals the average rate of change.

Applying these steps requires careful examination of function definitions and their derivatives.The function \(g(x)=x^{5/3}\) demonstrates these procedures perfectly:

• We verified its continuity and differentiability, applying MVT easily.
• The problem was solved by finding the specific \( c \) that satisfies the theorem.

Mastering these techniques in calculus problems enables solving more complex questions involving other functions and intervals.