To estimate the change in $f(x) = \tan^{-1}(x)$ from $x = a$ to $x = b$ using linear approximation, we will use the formula for linear approximation:

$f(b) \approx f(a) + f'(a)(b - a)$

Where:

• $f(x) = \tan^{-1}(x)$
• $a = 1$, $b = 1.04$

Step 1: Find $f(a)$ and $f'(a)$

The derivative of $f(x) = \tan^{-1}(x)$ is:

$f'(x) = \frac{1}{1 + x^2}$

At $x = a = 1$:

$f(a) = \tan^{-1}(1) = \frac{\pi}{4}$

And:

$f'(a) = \frac{1}{1 + 1^2} = \frac{1}{2}$

Step 2: Apply the Linear Approximation Formula

Using the formula:

$f(1.04) \approx \frac{\pi}{4} + \frac{1}{2}(1.04 - 1)$

Now, calculate this approximation.

$f(1.04) \approx 0.785398 + 0.5(0.04)$ $f(1.04) \approx 0.785398 + 0.02 = 0.805398$

So the linear approximation yields $f(1.04) \approx 0.805398$.

Step 3: Find the Actual Value of $f(1.04)$

We now calculate the actual value of $f(1.04) = \tan^{-1}(1.04)$:

$f(1.04) = \tan^{-1}(1.04) \approx 0.783324$

Step 4: Calculate the Error

The error is the difference between the actual value and the linear approximation:

$\text{Error} = | \text{Actual Value} - \text{Linear Approximation} |$ $\text{Error} = | 0.783324 - 0.805398 | \approx 0.022074$

Final Answer:

• Linear Approximation: $0.805398$
• Actual Value: $0.783324$
• Error: $0.022074$

Would you like further details or have any questions on the steps? Here are some related questions for exploration:

1. How can you improve the accuracy of the linear approximation method?
2. What is the relationship between the linear approximation and the second derivative of the function?
3. Can linear approximation be used for functions other than $\tan^{-1}(x)$? If so, how?
4. How does the error in linear approximation change with larger intervals?
5. What happens if we use a higher-order approximation like a Taylor series?

Tip: The linear approximation is most accurate when the interval between $a$ and $b$ is small. For larger changes in $x$, higher-order approximations (e.g., quadratic approximation) can give better results.