Problem 62 - Calculus Early Transcendentals

Question

Suppose \(f(x)=1 /(1+x)\) is to be approximated near \(x=0\). Find the linear approximation to \(f\) at 0 . Then complete the following table showing the errors in various approximations. Use a calculator to obtain the exact values. The percent error is \(100 |\) approximation \(-\) exact \(|/|\) exact \(| .\) Comment on the behavior of the errors as \(x\) approaches 0.

Step-by-Step Solution

Answer

Answer: As \(x\) approaches 0, the absolute error and the percent error for the linear approximation \(L(x) = 1 - x\) to the function \(f(x) = \frac{1}{1 + x}\) at \(x = 0\) decrease, indicating that the linear approximation becomes more accurate near the point of approximation.

Step 1: Evaluate the function at \(x = 0\)

To find the value of the function at \(x = 0\), we simply plug in \(x = 0\) into the function \(f(x) = \frac{1}{1 + x}\): $$ f(0) = \frac{1}{1 + 0} = 1 $$

Step 2: Find the first derivative of the function

Now let's find the derivative of function \(f(x) = \frac{1}{1 + x}\) with respect to \(x\). Using the power rule, we have: $$ f'(x) = -\frac{1}{(1 + x)^2} $$

Step 3: Apply the linear approximation formula

The formula for linear approximation is given by: $$ L(x) = f(a) + f'(a)(x - a) $$ Here, \(a = 0\). So, we can rewrite the formula as: $$ L(x) = f(0) + f'(0)(x - 0) $$ We know that \(f(0) = 1\), so now let's find \(f'(0)\): $$ f'(0) = -\frac{1}{(1 + 0)^2} = -1 $$ Thus, our linear approximation formula becomes: $$ L(x) = 1 - x $$

Step 4: Use a calculator to obtain the exact values, error, and percent error

In this step, we will use the linear approximation \(L(x) = 1 - x\) to find the approximate values for the given \(x\) values. Next, we will calculate the exact values for the same \(x\) values using the function \(f(x) = \frac{1}{1 + x}\). Afterward, we will find the absolute error and the percent error using the formulas: Absolute error \(= |\)approximate value \(-\) exact value\(|\) Percent error \(= \frac{100| \text{approximate value} - \text{exact value} |}{|\text{exact value}|}\) Here's the completed table: x L(x) f(x) Absolute Error Percent Error ---------------------------------------------------------------------- -0.1 1.1 1.1111 0.0111 1.000 -0.01 1.01 1.0101 0.0001 0.0100 -0.001 1.001 1.0010 0.0000 0.0010 0.1 0.9 0.9091 0.0091 1.000 0.01 0.99 0.9901 0.0001 0.0100 0.001 0.999 0.9990 0.0000 0.0010

Step 5: Comment on the behavior of the errors as x approaches 0

As we can see from the table, the error decreases as x approaches 0: from 1% error for x-values like -0.1 and 0.1 to 0.01% error for x-values like -0.01 and 0.01, and finally, 0.001% for x values like -0.001 and .001. This verifies that the linear approximation becomes more accurate near the point of approximation, which in this case is \(x=0\).

Key Concepts

Derivative

A derivative in mathematics measures how a function changes as its input changes. Think of it as capturing the idea of a function's "rate of change" — sort of like how fast something is moving at a certain point in time. In our exercise, we start with the function \(f(x) = \frac{1}{1+x}\) and find its derivative. To do this, we use a rule called the power rule. For our function, the derivative is \(f'(x) = -\frac{1}{(1+x)^2}\). This tells us how steep or flat the function is at any value of \(x\).
The derivative is crucial for linear approximation because it helps us understand how the function behaves around a specific point. Here, finding \(f'(0)\) gives us \(-1\), which is used in our linear approximation formula. This number indicates that for a tiny increase in \(x\), there's an approximate decrease of 1 in the function's value. By understanding derivatives, we can more easily predict how functions will behave near any given point.

Absolute Error

Absolute error helps us quantify the accuracy between an approximation and the exact value of a function. It's the difference between the approximate value obtained through methods like linear approximation and the true value calculated using the original function. In simple terms, it shows how much "off" our approximation can be.
Using the formula, Absolute error = |approximate value - exact value|, we can easily compute how accurate our linear approximation (e.g., \(L(x) = 1-x\)) is compared to the value from our exact function \(f(x) = \frac{1}{1 + x}\).
By looking at each value of \(x\), you subtract the linear approximation's result from the exact value. For example, when \(x = 0.1\), the linear approximation is \(0.9\), while the exact value is approximately \(0.9091\). The absolute error thus is the difference |0.9091 - 0.9|, which equals \(0.0091\). Comparing errors across various \(x\), we observe how close our approximation is to the actual function, highlighting its accuracy or lack thereof.

Percent Error

Percent error provides a way to put the error into perspective relative to the size of the exact value. It's particularly useful because it tells us how significant an error is in percentage terms, making it easier to understand and compare across different cases. The formula to calculate it is: \(\text{Percent error} = \frac{100 \times | \text{approximate value} - \text{exact value} | }{ | \text{exact value} | }\)Percent error is used to assess how good a linear approximation is. Smaller percent errors indicate a more accurate approximation.
For instance, at \(x = -0.1\), the percent error is calculated as \(1.000\%\). As \(x\) gets closer to zero, like for \(x = -0.001\), the percent error decreases significantly to \(0.001\%\). This shrinking error clearly demonstrates the effectiveness of our linear approximation near the point of linearization \(x = 0\). This trend reveals a crucial aspect of approximations: they generally work best when close to the point of expansion. Understanding percent errors allows us to make better judgments about the reliability of approximations in practical applications.