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Let's walk through the problem step by step to identify the mistakes and correct them. The task is to find the Taylor polynomial for \( f(x) = \sqrt{x + 3} \) about \( x = 1 \).

### Step 1: Calculating \( f(1) \)
The function is \( f(x) = \sqrt{x + 3} \). Plugging in \( x = 1 \):

\[
f(1) = \sqrt{1 + 3} = \sqrt{4} = 2
\]

So, \( f(1) = 2 \), which is correctly marked.

### Step 2: Calculating \( f'(x) \)
Using the chain rule, we differentiate:

\[
f'(x) = \frac{1}{2} (x + 3)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2} (x + 3)^{-\frac{1}{2}}
\]

Now, plug in \( x = 1 \):

\[
f'(1) = \frac{1}{2} (1 + 3)^{-\frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\]

This is correct, and marked correctly.

### Step 3: Calculating \( f''(x) \)
Differentiate \( f'(x) \) again:

\[
f''(x) = \frac{d}{dx} \left( \frac{1}{2} (x + 3)^{-\frac{1}{2}} \right)
\]
\[
f''(x) = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) (x + 3)^{-\frac{3}{2}} = -\frac{1}{4} (x + 3)^{-\frac{3}{2}}
\]

Now, plug in \( x = 1 \):

\[
f''(1) = -\frac{1}{4} (1 + 3)^{-\frac{3}{2}} = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32}
\]

This is correct, and marked correctly.

### Step 4: Calculating \( f'''(x) \)
Differentiate \( f''(x) \) again:

\[
f'''(x) = \frac{d}{dx} \left( -\frac{1}{4} (x + 3)^{-\frac{3}{2}} \right)
\]
\[
f'''(x) = -\frac{1}{4} \cdot \left( -\frac{3}{2} \right) (x + 3)^{-\frac{5}{2}} = \frac{3}{8} (x + 3)^{-\frac{5}{2}}
\]

Now, plug in \( x = 1 \):

\[
f'''(1) = \frac{3}{8} (1 + 3)^{-\frac{5}{2}} = \frac{3}{8} \cdot \frac{1}{32} = \frac{3}{256}
\]

This is also correct, and marked correctly.

### Step 5: Constructing the Taylor Polynomial
The formula for the Taylor polynomial of degree 3 is:

\[
T_3(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3
\]

Now, substitute the values:

\[
T_3(x) = 2 + \frac{1}{4}(x - 1) + \frac{-1}{32} \cdot \frac{1}{2}(x - 1)^2 + \frac{3}{256} \cdot \frac{1}{6}(x - 1)^3
\]
\[
T_3(x) = 2 + \frac{1}{4}(x - 1) - \frac{1}{64}(x - 1)^2 + \frac{1}{512}(x - 1)^3
\]

### Errors in the Expression
The last two terms in the Taylor polynomial are incorrect:
1. The coefficient for the \( (x - 1)^2 \) term should be \( -\frac{1}{64} \), but it's given as \( \frac{1}{512} \).
2. The \( (x - 1)^3 \) term coefficient is correct.

### Correct Taylor Polynomial:
\[
T_3(x) = 2 + \frac{1}{4}(x - 1) - \frac{1}{64}(x - 1)^2 + \frac{1}{512}(x - 1)^3
\]

Let me know if you need more details on any specific part!

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Here are 5 questions that expand on this:

1. How do you compute higher-order derivatives of \( f(x) = \sqrt{x+3} \)?
2. Why do we divide by \( n! \) when constructing Taylor polynomials?
3. How does the radius of convergence affect the accuracy of a Taylor polynomial?
4. How would the polynomial change if we expanded around a different point, say \( x = 0 \)?
5. Can you verify that the approximation for \( x = 1.5 \) using this Taylor polynomial is accurate?

**Tip:** When expanding Taylor series, always check the factorial in the denominator, as it affects higher-order terms significantly.

In this question, we will calculate the Taylor Polynomial for f(x) = sqrt(x + 3) about x = 1.

Step-by-step guide to finding the Taylor polynomial of degree 3 for f(x) = sqrt(x + 3) around x = 1. Includes detailed derivative calculations and correct polynomial approximation for calculus students.

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