Let's go through the process step by step.

1. Find the linear approximation of the function $f(x) = \sqrt{4 - x}$ at $a = 0$.

The linear approximation $L(x)$ of a function $f(x)$ at a point $a$ is given by the formula:

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

Step 1: Compute $f(a)$

Given that $f(x) = \sqrt{4 - x}$, let’s calculate $f(0)$:

$f(0) = \sqrt{4 - 0} = \sqrt{4} = 2$

Step 2: Compute $f'(x)$ (the derivative of $f(x)$)

Now, let's compute the derivative of $f(x)$:

$f'(x) = \frac{d}{dx} \left( \sqrt{4 - x} \right) = \frac{1}{2\sqrt{4 - x}} \cdot (-1) = \frac{-1}{2\sqrt{4 - x}}$

At $x = 0$:

$f'(0) = \frac{-1}{2\sqrt{4 - 0}} = \frac{-1}{2 \times 2} = \frac{-1}{4}$

Step 3: Write the linear approximation $L(x)$

Using the formula for linear approximation:

$L(x) = f(0) + f'(0)(x - 0) = 2 + \left( \frac{-1}{4} \right)x$

Thus, the linear approximation is:

$L(x) = 2 - \frac{1}{4}x$

2. Use $L(x)$ to approximate $\sqrt{3.9}$ and $\sqrt{3.99}$.

We need to approximate $\sqrt{3.9}$ and $\sqrt{3.99}$ using the function $f(x) = \sqrt{4 - x}$.

Approximate $\sqrt{3.9}$:

To approximate $\sqrt{3.9}$, observe that $3.9 = 4 - 0.1$, so we use $x = 0.1$ in $L(x)$:

$L(0.1) = 2 - \frac{1}{4}(0.1) = 2 - 0.025 = 1.975$

Thus, $\sqrt{3.9} \approx 1.975$.

Approximate $\sqrt{3.99}$:

To approximate $\sqrt{3.99}$, observe that $3.99 = 4 - 0.01$, so we use $x = 0.01$ in $L(x)$:

$L(0.01) = 2 - \frac{1}{4}(0.01) = 2 - 0.0025 = 1.9975$

Thus, $\sqrt{3.99} \approx 1.9975$.

Final answers:

$\sqrt{3.9} \approx 1.9750$ $\sqrt{3.99} \approx 1.9975$

Would you like more details on any step, or do you have any questions?

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Here are 5 related questions to deepen your understanding:

1. What is the purpose of a linear approximation?
2. How do we compute the derivative of more complex functions?
3. How accurate are linear approximations for values far from the point of approximation?
4. Can the method of linear approximation be applied to higher-degree functions?
5. What are the practical applications of linear approximations in physics or engineering?

Tip: The closer your value is to the approximation point, the more accurate your linear approximation will be.