Problem 15 - Calculus Single Variable

Question

Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=\ln (x), c=e^{3}, x=20 $$

Step-by-Step Solution

Answer

Using increments, we estimate \( f(20) \approx 2.996 \).

Step 1: Understand the Problem

We are given the function \( f(x) = \ln(x) \) and we need to estimate the value of this function at \( x = 20 \) using the known value \( f(c) \) where \( c = e^3 \). We will use the method of increments, which relies on linear approximations.

Step 2: Calculate the Increment in x

The increment in \( x \) is given by \( \Delta x = x - c \). Calculate \( \Delta x \) by substituting the given values of \( x = 20 \) and \( c = e^3 \). Substitute to get \( \Delta x = 20 - e^3 \). Since \( e^3 \) is approximately 20.0855, we find \( \Delta x \approx 20 - 20.0855 = -0.0855 \).

Step 3: Linear Approximation Formula

The linear approximation of \( f(x) \) near \( x = c \) is given by: \[ f(x) \approx f(c) + f'(c) \Delta x \]. We need to compute \( f(c) \) and \( f'(c) \).

Step 4: Compute f(c) and f'(c)

Since \( f(x) = \ln(x) \), then \( f'(x) = \frac{1}{x} \). Substitute \( c = e^3 \) to find: \[ f(c) = \ln(e^3) = 3 \] and \[ f'(c) = \frac{1}{e^3} \]. Note \( e^3 \approx 20.0855 \), so \( f'(c) \approx \frac{1}{20.0855} \approx 0.0497 \).

Step 5: Apply Linear Approximation

Substitute the values into the linear approximation formula: \[ f(x) \approx 3 + 0.0497 \times (-0.0855) \]. Calculate the approximation: \( 0.0497 \times (-0.0855) \approx -0.00424 \), therefore: \[ f(x) \approx 3 - 0.00424 = 2.99576 \].

Step 6: Conclusion

Thus, using the method of increments, we estimate \( f(20) \approx 2.99576 \).

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that focuses on the concept of the derivative. The derivative represents the rate at which a quantity changes. In simpler terms, it can be thought of as the "slope" of the function at any point.
It plays a vital role in finding linear approximations, as it provides the tangent's slope, which approximates the function locally.

When dealing with functions like natural logarithms, differential calculus allows us to find how the function behaves near a given point. For example, in our exercise, we have the function \( f(x) = \ln(x) \).
Its derivative, \( f'(x) = \frac{1}{x} \), is crucial for creating a linear approximation of the function.

• Differential calculus makes it easier to estimate function values near known points by providing a line tangent to the curve.
• It provides tools like derivatives, facilitating our approach to finding approximations.

Understanding differential calculus is essential in physics, engineering, and many scientific domains, as it leads to precise approximations and insights into function behavior.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a special logarithmic function that uses the number \( e \) as its base. \( e \) is an irrational and transcendental number approximately equal to 2.71828.
The natural logarithm has several unique properties:

• \( \ln(e) = 1 \) because \( e^1 = e \).
• \( \ln(e^a) = a \), simplifying logarithmic computations immensely.

In the context of our exercise, we use the property \( \ln(e^3) = 3 \) to find \( f(c) \) where \( c = e^3 \).
The function \( \ln(x) \) increases slowly, meaning small changes in \( x \) can lead to slight changes in the value of \( \ln(x) \).

This characteristic is useful in linear approximation, ensuring that the estimates remain close to the actual values.
Understanding natural logarithms is important in fields such as probability theory and complex numbers, providing a foundation for more advanced studies.

Method of Increments

The method of increments is a technique used in calculus to estimate the values of functions. This method leverages the idea of linear approximation by using small changes, or "increments," in \( x \), noted as \( \Delta x \).
In our exercise, we calculated \( \Delta x \) by finding the difference between the desired \( x \) value and the given \( c \) value: \( \Delta x = x - c \).

• This method simplifies calculations by employing the derivative \( f'(c) \).
• The formula \( f(x) \approx f(c) + f'(c) \Delta x \) helps us estimate the function's value.

It's essential in numerous applications where it is impractical to compute the function directly.
Such situations might include complex functions where a direct calculation could be overwhelming or impossible.

This approach remains a valuable tool in mathematical modeling, economics, and data sciences, helping to simplify predictions and understand changes in dynamic systems.