Problem 32 - Thomas Calculus

Question

In Exercises \(31-36,\) each function \(f(x)\) changes value when \(x\) changes from \(x_{0}\) to \(x_{0}+d x .\) Find a. the change \(\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)\) b. the value of the estimate \(d f=f^{\prime}\left(x_{0}\right) d x ;\) and c. the approximation error \(|\Delta f-d f|\) $$ f(x)=2 x^{2}+4 x-3, \quad x_{0}=-1, \quad d x=0.1 $$

Step-by-Step Solution

Answer

a. \(\Delta f = 0.02\), b. \(df = 0\), c. Approximation error is \(0.02\).

Step 1: Define the Function and Increment

The given function is \( f(x) = 2x^2 + 4x - 3 \). The initial point is \( x_0 = -1 \), and the increment is \( dx = 0.1 \). Our task is to compute the change in function value as \( x \) goes from \( x_0 \) to \( x_0 + dx \).

Step 2: Calculate \( f(x_0) \)

First, evaluate the function at \( x_0 = -1 \):\[ f(-1) = 2(-1)^2 + 4(-1) - 3 \]\[ f(-1) = 2(1) - 4 - 3 \]\[ f(-1) = 2 - 4 - 3 = -5 \]

Step 3: Calculate \( f(x_0 + dx) \)

Next, evaluate the function at \( x_0 + dx = -0.9 \):\[ f(-0.9) = 2(-0.9)^2 + 4(-0.9) - 3 \]\[ f(-0.9) = 2(0.81) - 3.6 - 3 \]\[ f(-0.9) = 1.62 - 3.6 - 3 = -4.98 \]

Step 4: Determine \( \Delta f \)

The change in the function value is the difference between \( f(x_0 + dx) \) and \( f(x_0) \):\[ \Delta f = f(-0.9) - f(-1) \]\[ \Delta f = -4.98 - (-5) \]\[ \Delta f = 0.02 \]

Step 5: Compute the Derivative \( f'(x) \)

Find the derivative of \( f(x) \), which is \( f'(x) = 4x + 4 \).

Step 6: Evaluate the Derivative at \( x_0 \)

Evaluate the derivative at \( x_0 = -1 \):\[ f'(-1) = 4(-1) + 4 \]\[ f'(-1) = -4 + 4 = 0 \]

Step 7: Calculate \( df \)

The estimate is given by \( df = f'(x_0) \cdot dx \):\[ df = 0 \cdot 0.1 = 0 \]

Step 8: Find the Approximation Error \( |\Delta f - df| \)

The error is the absolute value of the difference between \( \Delta f \) and \( df \):\[ |\Delta f - df| = |0.02 - 0| = 0.02 \]

Key Concepts

Function Change

When dealing with functions in calculus, understanding how a function changes as its input changes is key. This change is typically explored when you change the input value slightly, from an initial point. In our example, we use the function \( f(x) = 2x^2 + 4x - 3 \). Here, we check how it changes when \( x \) goes from an initial \( x_0 = -1 \) to \( x_0 + dx = -0.9 \):

• Start by calculating the function's value at the initial point: \( f(-1) = -5 \).
• Next, find the value at the new point: \( f(-0.9) = -4.98 \).
• The change in the function is \( \Delta f = f(-0.9) - f(-1) = 0.02 \).

This change, \( \Delta f \), reveals the nature and direction of the function's response to slight input shift, which is the essence of differential calculus.

Derivative Estimation

A cornerstone of differential calculus is using derivatives to estimate how much a function will change with a small input change. Derivative estimation allows for predicting this change without recalculating the function's values at two points. We use the derivative of the function \( f(x) = 2x^2 + 4x - 3 \), which is found by differentiating:

• The derivative, \( f'(x) = 4x + 4 \), offers a formula for the rate of change.
• At our initial point \( x_0 = -1 \), the derivative evaluates to \( f'(-1) = 0 \).
• Thus, the derivative estimate for our change \( df \) is: \( df = f'(x_0) \cdot dx = 0 \cdot 0.1 = 0 \).

This method gives a quick glimpse into the potential change and simplifies calculating slight variations over small intervals.

Approximation Error

In calculus, when we use derivatives to estimate function changes, we must consider the potential error in our approximation. The approximation error tells us how precise our derivative-based estimate is compared to the actual change. In our exercise:

• The actual change, \( \Delta f = 0.02 \), and the estimated change, \( df = 0 \), differ.
• The difference indicates the approximation error, calculated as: \[ |\Delta f - df| = |0.02 - 0| = 0.02 \].

This error highlights the limits of linear approximation, emphasizing that it's more accurate with smaller input changes or near the point where the derivative is calculated. Understanding this discrepancy helps guide us in understanding when and where derivative estimates serve as reliable proxies for function changes.