Problem 43 - 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) $$f(x)=x^{-1}, \quad x_{0}=0.5, \quad d x=0.1$$

Step-by-Step Solution

Answer

\( \Delta f = -\frac{1}{3} \), \( d f = -0.4 \), Error = \( \frac{1}{15} \).

Step 1: Calculate the Change \( \Delta f \)

To find the change in the function value, calculate \( \Delta f = f(x_{0} + dx) - f(x_{0}) \). Here, \( f(x) = x^{-1} \), \( x_{0} = 0.5 \), and \( dx = 0.1 \). First, find \( f(x_{0} + dx) = f(0.6) = (0.6)^{-1} = \frac{1}{0.6} \). Then, find \( f(x_{0}) = f(0.5) = (0.5)^{-1} = 2 \). Thus, \( \Delta f = \frac{1}{0.6} - 2 \).

Step 2: Simplify \( \Delta f \)

Further simplify \( \Delta f \):\[ \Delta f = \frac{1}{0.6} - 2 = \frac{5}{3} - 2 = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} \]This is the exact change in function value.

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

Calculate the derivative of \( f(x) = x^{-1} \). Using the power rule, we find \( f'(x) = -x^{-2} \). Substitute \( x_0 = 0.5 \) to find \( f'(0.5) = -(0.5)^{-2} = -4 \).

Step 4: Calculate the Estimate \( d f \)

Use the derivative to estimate the change: \( d f = f'(x_0) \cdot dx \). With \( f'(0.5) = -4 \) and \( dx = 0.1 \), compute \( d f = (-4) \cdot 0.1 = -0.4 \).

Step 5: Calculate the Approximation Error

The approximation error is \( |\Delta f - d f| \). Substitute \( \Delta f = -\frac{1}{3} \) and \( d f = -0.4 \):\[ |\Delta f - d f| = \left| -\frac{1}{3} + 0.4 \right| = \left| -\frac{1}{3} + \frac{2}{5} \right| \]Convert terms to a common denominator:\[ |\Delta f - d f| = \left| -\frac{5}{15} + \frac{6}{15} \right| = \frac{1}{15} \].

Key Concepts

Function Change

Understanding how a function changes as its input changes is a core idea in calculus. When we talk about the change in a function like \( f(x) \) when \( x \) changes, we're discussing what's called \( \Delta f \). This represents the difference between the function's value at two different points.

• \( \Delta f = f(x_0 + dx) - f(x_0) \)
• It's the actual change of the function as \( x \) moves a small amount \( dx \).

In our example, we see how this change is calculated for \( f(x) = x^{-1} \). First, we find \( f(0.6) = (0.6)^{-1} = \frac{1}{0.6} \), and similarly, \( f(0.5) = 2 \). The change \( \Delta f \) is then the difference \( \frac{1}{0.6} - 2 = -\frac{1}{3} \). This illustrates the exact shift in the function's value due to the increase in \( x \).

Derivative Estimation

The derivative is a powerful tool in calculus. It allows us to estimate how a function behaves as its input changes. Specifically, it gives us a good approximation of the function's rate of change around a point.

• The derivative of \( f(x) = x^{-1} \) is \( f'(x) = -x^{-2} \).
• By evaluating the derivative at \( x_0 = 0.5 \), we find \( f'(0.5) = -4 \).

We use this derivative to estimate the change in the function value over a small interval. The estimate of the change, denoted as \( df \), is calculated by \( df = f'(x_0) \cdot dx \). In our scenario, with \( dx = 0.1 \), this gives \( df = (-4) \cdot 0.1 = -0.4 \). Although a close approximation, this method provides an estimation, not the exact change.

Approximation Error

Even with the best mathematical tools, sometimes our estimates will differ from actual values. This difference is known as the approximation error. Recognizing and calculating this error helps understand how closely an estimate aligns with reality.Consider the expression for approximation error,

• \( |\Delta f - df| \) quantifies how far off the estimated change (\( df \)) is from the actual change (\( \Delta f \)).

In our problem, we find \(|\Delta f - df| = \left|-\frac{1}{3} + 0.4\right| = \frac{1}{15} \). This shows that the derivative's approximation, while useful, is not perfect. Understanding these errors is vital, especially in more complex calculations or where precise accuracy is necessary.Always keep in mind the impact of such errors, particularly when making decisions or predictions based on your calculus analysis.