We are asked to find the linear approximation $L(x)$ of the function $f(x) = \sqrt{x + 3}$ at $a = 22$.

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

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

Step 1: Calculate $f(a)$

Given $f(x) = \sqrt{x + 3}$, we can substitute $a = 22$:

$f(22) = \sqrt{22 + 3} = \sqrt{25} = 5$

Step 2: Find the derivative of $f(x)$

The derivative of $f(x) = \sqrt{x + 3}$ is:

$f'(x) = \frac{d}{dx} \left( (x + 3)^{1/2} \right) = \frac{1}{2} (x + 3)^{-1/2} = \frac{1}{2\sqrt{x + 3}}$

Step 3: Calculate $f'(a)$

Now, we substitute $a = 22$ into the derivative:

$f'(22) = \frac{1}{2 \sqrt{22 + 3}} = \frac{1}{2 \sqrt{25}} = \frac{1}{2 \times 5} = \frac{1}{10}$

Step 4: Write the linear approximation

Now, we substitute into the linear approximation formula:

$L(x) = f(22) + f'(22)(x - 22)$ $L(x) = 5 + \frac{1}{10}(x - 22)$

Distribute the $\frac{1}{10}$:

$L(x) = 5 + \frac{1}{10}x - \frac{22}{10}$

Simplify the constant term:

$L(x) = \frac{5}{1} - \frac{22}{10} = \frac{50}{10} - \frac{22}{10} = \frac{28}{10} = 2.8$

Thus, the linear approximation is:

$L(x) = \frac{1}{10}x + 2.8$

This corresponds to the second option in the image:

$\boxed{L(x) = \frac{1}{10}x + \frac{14}{10}}$

Would you like more details or have any other questions? Here are some related questions:

1. How is linear approximation different from a Taylor series?
2. Can we use linear approximation to estimate non-integer values of the function?
3. How do we decide which point $a$ to use for approximation?
4. What other techniques can be used to approximate functions?
5. Why does the derivative play a crucial role in linear approximations?

Tip: Always simplify the constants when performing linear approximations to make your result more interpretable.