Problem 66 - Essential Calculus

Question

In Exercises 65 and 66, apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) \(\begin{array}{l} \underline{\text { Functions }} \\ f(x)=\ln x, \quad g(x)=x^{3} \end{array} \quad \frac{\text { Interval }}{\left[1,4\right]}\)

Step-by-Step Solution

Answer

The value \( c \) within the interval [1,4] that satisfies the extension of Mean Value Theorem for the functions \( f(x) = \ln x \) and \( g(x) = x^3 \) is \( c = \sqrt[3]{\frac{63}{3 \ln 4}} \).

Step 1: Understanding the Expression

Here, we are given two functions \( f(x) = \ln x \) and \( g(x) = x^3 \) and an interval [1,4]. The expression \( \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \) is the Extended Mean Value Theorem.

Step 2: Compute the Derivatives of the Functions

The derivative of \( f(x) = \ln x \) is \( f'(x) = \frac{1}{x} \). The derivative of \( g(x) = x^3 \) is \( g'(x) = 3x^2 \).

Step 3: Substitute the Given Functions into the Expression

Set up the expression to find \( c \). \nSo, \(\frac{f'(c)}{g'(c)} = \frac{f(4) - f(1)}{g(4) - g(1)} \) becomes \( \frac{\frac{1}{c}}{3c^2} = \frac{\ln 4 - \ln 1}{4^3 - 1^3} \). Simplifying right side we have \( \frac{\ln 4}{63} \).

Step 4: Simplify and Solve for \( c \)

Cross multiply and solve for \( c \): \( \frac{1}{c} = \frac{3c^2 \ln 4}{63} \). Solving this equation for \( c \), we get \( c^3 = \frac{63}{3 \ln 4} \), so \( c = \sqrt[3]{\frac{63}{3 \ln 4}} \). We can confirm that this solution lies within the interval [1,4].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique particularly useful when dealing with functions raised to a power or products and quotients of functions. Here, it's applied to the function \( f(x) = \ln x \). By taking the natural logarithm of both sides of a more complex equation, you can simplify the differentiation process, especially for problems involving exponential functions or products. In the context of our problem, however, the function is already expressed simply as a natural logarithm, so the differentiation is straightforward.

• This technique helps convert multiplicative complexities into additive ones.
• It's beneficial for functions that are difficult to differentiate directly.
• It often simplifies the process when dealing with exponentials or polynomials.

Understanding logarithmic differentiation ensures you are prepared for more complex problems and helps in simplifying your work when the typical differentiation rules become cumbersome. Remember, good comprehension of basic logarithmic properties and differentiation rules is crucial for successfully applying this method.

Polynomial Functions

Polynomial functions are among the most simple and frequently encountered mathematical expressions. In our exercise, \( g(x) = x^3 \) is a polynomial function. This type of expression is characterized by consisting of variables raised to whole-number powers. The simplicity of polynomials makes them important in calculus because they possess derivatives of every order.
Whenever differentiating polynomial functions, each term is treated independently, and the power rule is typically applied: differentiate \( x^n \) as \( nx^{n-1} \).
Here are a few basic points about polynomial functions:

• They provide a foundational element in calculus study.
• Polynomial functions are continuous and differentiable.
• They are useful for approximating more complex functions.

The simplicity of polynomial functions like \(g(x) = x^3\) allows us to focus on using their straightforward derivatives for applications like computing rates of change or finding tangents, which are critical for understanding concepts such as the Extended Mean Value Theorem.

Derivative Calculations

Calculating derivatives is a core task in calculus and is essential for understanding how functions change. In the provided solution, we find derivatives of \( f(x) = \ln x \) and \( g(x) = x^3 \). For \( f(x) = \ln x \), its derivative is \( f'(x) = \frac{1}{x} \), using the fundamental derivative of the natural logarithm function. For \( g(x) = x^3 \), we apply the power rule to find \( g'(x) = 3x^2 \).
The correctness of derivative calculation can be verified by checking:

• Use differentiation rules accurately (e.g., power rule, chain rule).
• Simplify the derivatives correctly to make the calculations manageable.
• Substitute the derivatives back into the problem to validate their necessity.

In the case of our exercise related to Extended Mean Value Theorem, precise calculation and substitution of derivatives allowed us to solve for the value \( c \), ensuring it lies within the interval [1,4]. Accurate derivative calculations lay the groundwork for more complex calculus applications, enabling a deeper understanding of the behavior and properties of functions over an interval.