Problem 21 - 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 \(f(c)\) $$ f(x)=\sqrt{2} \cos (\pi /(x+3)), c=1, x=0.81 $$

Step-by-Step Solution

Answer

Using increments, estimated value of \( f(0.81) \) is approximately 0.9627.

Step 1: Determine the increment \( \Delta x \)

The increment, \( \Delta x \), is the difference between \( x \) and \( c \). Calculate this by subtracting \( c \) from \( x \). \[ \Delta x = x - c = 0.81 - 1 = -0.19 \]

Step 2: Evaluate \( f(c) \)

Substitute \( x = c = 1 \) into the function \( f(x) \). \[ f(1) = \sqrt{2} \cos(\pi/(1+3)) = \sqrt{2} \cos(\pi/4) = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1 \]

Step 3: Find the derivative \( f'(x) \)

Calculate the derivative of \( f(x) = \sqrt{2} \cos(\pi/(x+3)) \) with respect to \( x \). Use the chain rule. \[ f'(x) = -\sqrt{2} \cdot \sin(\pi/(x+3)) \cdot \left(-\frac{\pi}{(x+3)^2}\right) = \frac{\sqrt{2} \pi \sin(\pi/(x+3))}{(x+3)^2} \]

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

Substitute \( c = 1 \) into the derivative \( f'(x) \) to find \( f'(1) \). \[ f'(1) = \frac{\sqrt{2} \pi \sin(\pi/4)}{(1+3)^2} = \frac{\sqrt{2} \cdot \pi \cdot \frac{\sqrt{2}}{2}}{16} = \frac{\pi}{16} \]

Step 5: Calculate the approximate value of \( f(x) \)

Use the method of increments to estimate \( f(x) \) by applying the formula \( f(x) \approx f(c) + f'(c) \Delta x \). \[ f(0.81) \approx 1 + \frac{\pi}{16} \cdot (-0.19) = 1 - \frac{0.19 \pi}{16} \] Evaluating numerically, \[ f(0.81) \approx 1 - 0.0373 \approx 0.9627 \]

Key Concepts

Method of Increments

The Method of Increments is a fundamental idea used to approximate function values. This process involves small changes in a function’s input and observing the resultant change in the output.
This technique is particularly useful when you have known values and derivatives that can help estimate unknown function values nearby. To break it down further:

• We begin by identifying an increment, denoted as \( \Delta x \). This is simply the difference between your point of estimation, \( x \), and a nearby point \( c \) where the function's value is known.
• Next, using this increment, a linear approximation formula is applied: \( f(x) \approx f(c) + f'(c) \Delta x \). This formula adds the change in output due to \( \Delta x \) to the known output \( f(c) \).
• The derivative, \( f'(c) \), acts as a measure of the function's sensitivity to changes in input around \( c \).

This method is quite efficient for estimation, especially when you cannot directly compute \( f(x) \). By leveraging both the function's behavior at \( c \) and its rate of change, the method of increments offers a pragmatic approach to function approximation.

Derivative Calculation

The derivative calculation is crucial as it determines the function’s rate of change. This step allows us to express how a small change in \( x \) produces a change in \( f(x) \).
In our problem, we used the chain rule to find the derivative of the function \( f(x) = \sqrt{2} \cos(\pi/(x+3)) \). The chain rule helps when dealing with composite functions.

• First, identify the outer function and the inner function. Here, \( \cos \) is the outer function while \( \pi/(x+3) \) is the inner one.
• Begin by deriving the outer function with respect to the inner one, i.e., \( (\cos u)' = -\sin u \).
• Then multiply the result by the derivative of the inner function, \( (\pi/(x+3))' = -\frac{\pi}{(x+3)^2} \).
• The final derivative expression becomes: \( f'(x) = \frac{\sqrt{2} \pi \sin(\pi/(x+3))}{(x+3)^2} \).

This derivative gives us insight into how sensitive the function is around point \( c \), thus playing a vital role in the method of increments.

Function Approximation

Function approximation is about estimating unknown function values using known data and derivative information. It leverages linear approximations to deliver close estimates of function values.
Using information from a familiar point (\( c \)), and how the function behaves near this point (via its derivative), we can make educated guesses for \( f(x) \). Here's how it’s typically performed:

• Start with a known value, \( f(c) \), which acts as a base for the approximation.
• Utilize the derivative at this point, \( f'(c) \), to understand the function’s tendency—whether it is increasing or decreasing.
• Apply the formula \( f(x) \approx f(c) + f'(c) \Delta x \), which combines the known value and the product of the derivative with the increment to predict \( f(x) \).
• This linear combination provides a reasonable estimate of the target function value, \( f(x) \), using tangential insights.

This technique underlies many numerical methods and is foundational in calculus where precise solutions may be difficult to compute directly.