Problem 40 - University Calculus: Early Transcendentals

Question

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 \(|\Delta f-d f|\) (GRAPH CANNOT COPY) $$f(x)=2 x^{2}+4 x-3, x_{0}=-1, \quad d x=0.1$$

Step-by-Step Solution

Answer

\( \Delta f = 0.02 \), \( df = 0 \), Approximation error = 0.02.

Step 1: Calculate the Change in Function Value

We need to find the change \( \Delta f \) in the function value as \( x \) goes from \( x_0 = -1 \) to \( x_0 + dx = -0.9 \). The change \( \Delta f \) is given by:\[\Delta f = f(x_0 + dx) - f(x_0)\]Substitute in the values:\[ f(x_0 + dx) = f(-0.9) = 2(-0.9)^2 + 4(-0.9) - 3 \]Calculate \( f(-0.9) \):\[ = 2(0.81) - 3.6 - 3 = 1.62 - 3.6 - 3 = -4.98 \]Next, calculate \( f(x_0) \):\[ f(-1) = 2(-1)^2 + 4(-1) - 3 = 2 - 4 - 3 = -5 \]Now, \( \Delta f \) is:\[ \Delta f = f(-0.9) - f(-1) = -4.98 - (-5) = 0.02 \]

Step 2: Calculate the Estimate Using the Derivative

To estimate \( df \), we use the derivative of \( f(x) \) at \( x_0 = -1 \). The derivative \( f'(x) \) is:\[ f'(x) = \frac{d}{dx}(2x^2 + 4x - 3) = 4x + 4 \]Evaluate the derivative at \( x_0 = -1 \):\[ f'(-1) = 4(-1) + 4 = -4 + 4 = 0 \]The estimated change \( df \) is:\[ df = f'(-1) \cdot dx = 0 \cdot 0.1 = 0 \]

Step 3: Calculate the Approximation Error

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

Step 4: Summarize Calculations

To summarize:a. The change \( \Delta f = 0.02 \).b. The value of the estimate \( df = 0 \).c. The approximation error \( |\Delta f - df| = 0.02 \).

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus. They measure how a function changes as its input changes.A derivative is denoted by \( f'(x) \), and it is essentially the slope of the tangent line to the function at any given point. This indicates the rate at which the function is changing.
Think of it like driving a car: the derivative tells you how fast you are going (the speed) at a specific point in time. For our function \( f(x) = 2x^2 + 4x - 3 \), the derivative \( f'(x) \) is calculated as:

• The derivative of \( 2x^2 \) is \( 4x \)
• The derivative of \( 4x \) is \( 4 \)
• The derivative of a constant like \(-3\) is \( 0 \)

Thus, the derivative becomes \( f'(x) = 4x + 4 \). This tells us how \( f(x) \) is changing in response to small changes in \( x \). Evaluating this at \( x_0 = -1 \), we find \( f'(-1) = 0 \). This means at \( x_0 = -1 \), the rate of change is \( 0 \), indicating no change in function value at this point.

Function Approximation

Function approximation is a valuable technique in calculus, allowing us to estimate function values at certain points without calculating their exact values. When we have a tiny change in \( x \), given by \( dx \), the derivative helps us estimate how much the function \( f(x) \) will change. This estimate is denoted as \( df \), using the formula \( df = f'(x_0) \cdot dx \).
For our case, with \( f(x) = 2x^2 + 4x - 3 \), we use the derivative \( f'(-1) = 0 \). Hence, the estimated change \( df \) becomes \( 0 \cdot 0.1 = 0 \). This shows that, according to the linear approximation, the value of the function should not change when moving from \( x_0 = -1 \) to \( x_0 + dx = -0.9 \). Keep in mind, this is just an approximation – it may or may not be perfect.
The beauty of function approximation is that it simplifies complex calculations involving functions in many real-world applications.

Error Analysis in Approximation

Error analysis is important as it helps us understand the accuracy of our estimates and how closely they reflect reality. When estimating a function's change using its derivative, we compare the true change \( \Delta f \) and the estimated change \( df \).
In this exercise, the true change \( \Delta f \) was calculated as \( 0.02 \), while the estimated change \( df \) was \( 0 \). The approximation error, therefore, is given by \( |\Delta f - df| \), which equals \( |0.02 - 0| = 0.02 \).
This error signifies the difference between what actually happened (the function's change) and what we estimated using the derivative. Such analysis is key in refining our models and methods for improvement. Errors in approximation highlights factors not considered, such as curvature of the original function in this case. By understanding the deviation between our approximation and the actual result, we can better estimate and arrive at more accurate predictions in practical scenarios.