https://file.aiquickdraw.com/mathbot/file/1728145940476i08uej6h.jpg

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.

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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.

Consider the function f(x) = sqrt(x + 99). Find the linearization of f at a = 1, and use it to approximate sqrt(99), rounded to six decimal places.

Learn how to linearize the function f(x) = sqrt(x + 99) at x = 1 and use it to approximate sqrt(99) to six decimal places. This step-by-step solution explores key calculus concepts, such as linearization and derivative approximation.

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