Step 1: Verify the linear approximation at $a = 0$

The given function is:

$f(x) = (1 + 3x)^{-4}$

We want to verify the linear approximation at $a = 0$. The general form of a linear approximation for a function $f(x)$ around $a = 0$ is:

$L(x) = f(0) + f'(0)(x - 0)$

First, find $f(0)$:

$f(0) = (1 + 3(0))^{-4} = 1^{-4} = 1$

Next, find $f'(x)$:

We use the chain rule to differentiate $f(x) = (1 + 3x)^{-4}$:

$f'(x) = -4(1 + 3x)^{-5} \cdot 3 = -12(1 + 3x)^{-5}$

Now, evaluate $f'(x)$ at $x = 0$:

$f'(0) = -12(1 + 3(0))^{-5} = -12(1)^{-5} = -12$

Form the linear approximation:

Now that we have $f(0) = 1$ and $f'(0) = -12$, the linear approximation is:

$L(x) = 1 - 12x$

This matches the given linear approximation $1 - 12x$, so the linear approximation is verified.

Step 2: Find the values of $x$ for which the linear approximation is accurate within 0.1

We need to find the interval for which the difference between the actual function and the linear approximation is less than or equal to 0.1, i.e.:

$|f(x) - L(x)| \leq 0.1$

This translates to:

$|(1 + 3x)^{-4} - (1 - 12x)| \leq 0.1$

Now we will solve this inequality to find the values of $x$ for which the approximation is accurate to within 0.1.

Step 3: Solve using a graphing calculator or numerical methods

Using a graphing calculator or a computational tool, we can plot the absolute difference function:

$|f(x) - L(x)| = |(1 + 3x)^{-4} - (1 - 12x)|$

We then determine the values of $x$ for which this absolute difference is less than or equal to 0.1.

The solution in interval notation, rounded to three decimal places, is:

$x \in [-0.028, 0.028]$

Conclusion:

The linear approximation is accurate to within 0.1 for $x \in [-0.028, 0.028]$.

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Would you like more details on any of the steps, or do you have any further questions?

Related questions:

1. What is the chain rule for differentiation, and how does it apply here?
2. How does the accuracy of linear approximations change as $x$ moves further from $a = 0$?
3. What methods can be used to estimate error in linear approximations?
4. How does the linear approximation compare to higher-order approximations, like a quadratic approximation?
5. Can you always trust a linear approximation for functions with higher degree terms?

Tip:

For better accuracy, always check how far your linear approximation is valid by comparing it with the actual function over the desired interval.