Let's solve the problem in two parts, as outlined in the image:

Part (a): Find the linearization $L(x)$ of the function $f(x) = \sqrt{x + 99}$ at $a = 1$.

The linearization of a function $f(x)$ at $a = 1$ is given by the formula: $L(x) = f(a) + f'(a)(x - a)$

Step 1: Calculate $f(a)$

Since $a = 1$ and $f(x) = \sqrt{x + 99}$, we calculate $f(1)$: $f(1) = \sqrt{1 + 99} = \sqrt{100} = 10$

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

Using the chain rule, the derivative of $f(x) = \sqrt{x + 99}$ is: $f'(x) = \frac{1}{2\sqrt{x + 99}}$

Step 3: Evaluate $f'(a)$

Now, substitute $a = 1$ into $f'(x)$: $f'(1) = \frac{1}{2\sqrt{1 + 99}} = \frac{1}{2\sqrt{100}} = \frac{1}{20}$

Step 4: Write the linearization $L(x)$

Substitute the values of $f(1)$ and $f'(1)$ into the linearization formula: $L(x) = 10 + \frac{1}{20}(x - 1)$ So the linearization is: $L(x) = 10 + \frac{1}{20}(x - 1)$

Part (b): Use the linearization to approximate $\sqrt{99}$

We want to approximate $\sqrt{99}$ using the linearization. Notice that $\sqrt{99}$ is $f(0)$ because $f(x) = \sqrt{x + 99}$, and if $x = 0$, then $f(0) = \sqrt{99}$.

So, we substitute $x = 0$ into the linearization $L(x)$: $L(0) = 10 + \frac{1}{20}(0 - 1) = 10 + \frac{1}{20}(-1) = 10 - \frac{1}{20} = 10 - 0.05 = 9.95$

Thus, using the linearization, the approximation for $\sqrt{99}$ is: $\sqrt{99} \approx 9.950000$

Final Answer:

• (a) The linearization is $L(x) = 10 + \frac{1}{20}(x - 1)$.
• (b) The approximation for $\sqrt{99}$ is approximately $9.950000$, rounded to six decimal places.

Would you like more details on any part of this solution, or do you have any further questions?

Here are 5 related questions to explore:

1. How does linearization provide an approximation for functions near a point?
2. What is the general formula for the linearization of a function?
3. How accurate is the approximation of $\sqrt{99}$ compared to the actual value?
4. Can linearization be used for any type of function?
5. What are the limitations of using linearization for approximations?

Tip: The accuracy of linearization depends on how close $x$ is to the point of linearization $a$. The closer $x$ is to $a$, the more accurate the approximation.