Problem 35 - 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 \(|\Delta f-d f|\) $$ f(x)=x^{-1}, \quad x_{0}=0.5, \quad d x=0.1 $$

Step-by-Step Solution

Answer

The change is \(-0.3333\), the estimate is \(-0.4\), and the error is \(0.0667\).

Step 1: Compute Δf

First, calculate the function values at \(x_0 + dx\) and \(x_0\): \[ f(x) = x^{-1} \implies f(x_0 + dx) = (x_0 + dx)^{-1} \] Given \(x_0 = 0.5\) and \(dx = 0.1\), we have: \[ f(0.5) = (0.5)^{-1} = 2 \] \[ f(0.5 + 0.1) = (0.6)^{-1} = \frac{1}{0.6} = \frac{5}{3} \approx 1.6667 \] Thus, \[ \Delta f = f\left(x_0 + dx\right) - f\left(x_0\right) = \frac{5}{3} - 2 = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} \approx -0.3333 \]

Step 2: Compute df

Find the derivative of the function \(f(x) = x^{-1}\): \[ f'(x) = -x^{-2} \] Substitute \(x_0 = 0.5\) into the derivative: \[ f'(0.5) = -(0.5)^{-2} = -4 \] Multiply by \(dx = 0.1\) to find \(df\): \[ df = f'(0.5) \cdot dx = -4 \times 0.1 = -0.4 \]

Step 3: Compute Approximation Error

Calculate the approximation error as the absolute value of the difference between \(\Delta f\) and \(df\): \[ |\Delta f - df| = |-0.3333 - (-0.4)| = |-0.3333 + 0.4| = |0.0667| = 0.0667 \]

Key Concepts

Function Approximation

In calculus, function approximation is the process of finding a simpler expression that closely matches the values of a more complex function over a specific range. This concept is fundamental when analyzing functions because calculating precise values can be difficult or impossible. Function approximation allows us to make reasonably accurate predictions even without having the exact function details available. We use various techniques depending on the math scenario, but one common method found in differential calculus is using derivatives.

For instance, the original exercise involves a function, specifically \( f(x) = x^{-1} \). Our goal is to approximate how this function changes as \( x \) shifts from \( x_0 = 0.5 \) to \( x_0 + dx = 0.6 \). With approximation, we can predict function behavior more easily, especially when dealing with complex curves or when exact computation is impractical.

Function approximation is not only useful in academic exercises. It's widely used in real-world applications including computer graphics, economics, and scientific computations to model behavior in simpler terms.

Derivative

The derivative is a core concept in calculus that represents the rate of change of a function. It shows us how the function's output alters as its input changes. In practical terms, it's like determining the slope at any point on the curve of the function.

In our exercise, we start with the function \( f(x) = x^{-1} \). The derivative \( f'(x) \) helps to estimate how much the function is going to change for a tiny shift in \( x \). We find that the derivative is \( f'(x) = -x^{-2} \). This derivative indicates how steeply the curve slopes and how rapid the change is at any point.

By calculating \( f'(0.5) = -4 \), we're saying that near \( x = 0.5 \), for every small change in \( x \), \( f(x) \) will decrease quite sharply. Thus, using a derivative, we construct a linear approximation which can be quite close to the actual function value.

Approximation Error

Approximation error measures the difference between the estimated value of a function and its actual value. This is essential to know because no approximation is ever perfect, and understanding its error helps in determining the reliability of our approximation.

In our example, we first compute \( \Delta f \), which represents the actual change in function value as \( x \) moves from 0.5 to 0.6. This gives us a change of \( -\frac{1}{3} \). However, when using the derivative method, we get \( df = -0.4 \) as our estimate.

The approximation error, therefore, is the absolute value of \( |\Delta f - df| = |0.0667| \). This small error shows that the derivative-based approximation (linear approximation) is reasonably close to the actual change. Recognizing and calculating such errors is vital in fields that rely heavily on numerical methods, such as engineering and physics, where precision is crucial.