Problem 19 - 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 \(f(c)\) $$ f(x)=16^{x}, c=0, x=0.13 $$

Step-by-Step Solution

Answer

The approximate value of \( f(0.13) \) is 1.3604.

Step 1: Understand the function and given points

We have the function \( f(x) = 16^x \), and we need to estimate \( f(x) \) at \( x = 0.13 \) given \( f(c) \) where \( c = 0 \). Thus, \( f(c) = f(0) = 16^0 = 1 \).

Step 2: Calculate the increment \(
x \)

The increment in \( x \) from \( c \) to \( x \) is given by \( \Delta x = x - c = 0.13 - 0 = 0.13 \).

Step 3: Differentiate the function to find \( f'(x) \)

First, express \( f(x) = 16^x \) as an exponential function: \( f(x) = e^{x \ln 16} \). Then, the derivative is \( f'(x) = \ln(16) \cdot 16^x \).

Step 4: Calculate \( f'(c) \)

Substitute \( c = 0 \) into the derivative to find \( f'(0) \): \[ f'(0) = \ln(16) \times 16^0 = \ln(16) \times 1 = \ln(16). \]

Step 5: Apply the method of increments

The method of increments approximates \( f(x) \) as: \[ f(x) \approx f(c) + f'(c) \cdot \Delta x. \] Substitute the known values: \[ f(0.13) \approx 1 + \ln(16) \times 0.13. \]

Step 6: Compute the approximation

Calculate \( \ln(16) \) (which is about 2.7726), and then \( 1 + 2.7726 \times 0.13 \), which computes to approximately \( 1 + 0.3604 = 1.3604 \).

Key Concepts

Exponential Functions

Exponential functions are fascinating and vital in mathematics, especially in calculus and applied sciences. An exponential function expresses a mathematical model where a constant base is raised to a variable exponent.
For instance, in our exercise, the function is defined as \( f(x) = 16^x \), where the base is 16 and the exponent is \( x \). This means that as the exponent increases, the function grows rapidly.
An important characteristic of exponential functions is their growth or decay rate, which can be extraordinarily fast, depending on the base.

• If the base is greater than 1, like in \( 16^x \), the function grows.
• If the base is between 0 and 1, the function decays.

Understanding exponential functions allows us to model and predict various phenomena, such as population growth, radioactive decay, and even financial investments.

Derivative Calculation

Derivative calculation is an essential concept in calculus, helping us to understand how functions change. The derivative of a function indicates the rate at which the function's value changes at a certain point.
For functions like \( 16^x \), which can be expressed as \( e^{x \ln 16} \), the derivative \( f'(x) \) involves applying the chain rule.
Here’s a breakdown of the calculation:

• Re-express \( f(x) \) using the natural exponential function: \( f(x) = e^{x \ln 16} \).
• Differentiate using the chain rule: \( f'(x) = \ln(16) \cdot 16^x \).

Calculating derivatives provides insight into the behavior of functions and is foundational in understanding curves and slopes.

Increment Approximation

Using increment approximation, we can estimate the value of a function close to a known point. This method leverages small changes in x, known as increments, to find the function’s approximate value at a nearby point.
The "Method of Increments" involves:

• Identifying \( \Delta x \), the change in \( x \), which is \( 0.13 \) in our example.
• Using the derivative at a known point \( c \), which is \( f'(c) = \ln(16) \).
• Applying the formula: \( f(x) \approx f(c) + f'(c) \cdot \Delta x \).

This technique, despite its simplicity, is a powerful approach to quickly estimating values without complex calculations.

Calculus Estimation Methods

Calculus provides various methods for estimating functions' values, each with its own advantages. The "Method of Increments" is just one of these efficient techniques.
Its strength lies in:

• Simplifying complex functions into more manageable linear approximations.
• Offering quick, accurate estimates for small changes in \( x \).
• Reducing computation time compared to exact methods.

By understanding and using these methods, students can solve problems involving real-world phenomena faster and with reasonable accuracy. Such estimation techniques are crucial in fields such as engineering, economics, and even computer science.