ind the Taylor polynomial of order 3 generated by $$ f(x)=\ln \sqrt{x} $$ at the point $$ a=2 $$

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To find the Taylor polynomial of order 3 for the function $$ f(x) = \ln \sqrt{x} $$ centered at $$ a = 2 $$, follow these steps:

### 1. Simplify the Function
First, simplify $$ f(x) $$:
$$
f(x) = \ln \sqrt{x} = \frac{1}{2} \ln x
$$

### 2. Compute the Derivatives
Next, compute the derivatives of $$ f(x) $$ up to the third order:

- **First derivative:**
$$
f'(x) = \frac{d}{dx}\left(\frac{1}{2} \ln x\right) = \frac{1}{2x}
$$
$$
f'(2) = \frac{1}{2 \times 2} = \frac{1}{4}
$$

- **Second derivative:**
$$
f''(x) = \frac{d}{dx}\left(\frac{1}{2x}\right) = -\frac{1}{2x^2}
$$
$$
f''(2) = -\frac{1}{2 \times 4} = -\frac{1}{8}
$$

- **Third derivative:**
$$
f'''(x) = \frac{d}{dx}\left(-\frac{1}{2x^2}\right) = \frac{1}{x^3}
$$
$$
f'''(2) = \frac{1}{8}
$$

### 3. Construct the Taylor Polynomial
The Taylor polynomial of degree 3 centered at $$ a = 2 $$ is given by:
$$
P_3(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f'''(2)}{3!}(x - 2)^3
$$

Plugging in the computed values:
$$
P_3(x) = \frac{1}{2} \ln 2 + \frac{1}{4}(x - 2) - \frac{1}{16}(x - 2)^2 + \frac{1}{48}(x - 2)^3
$$

### Final Answer
$$
P_3(x) = \frac{1}{2} \ln 2 + \frac{1}{4}(x - 2) - \frac{1}{16}(x - 2)^2 + \frac{1}{48}(x - 2)^3
$$
The Taylor polynomial of order 3 for $$ f(x) = \ln \sqrt{x} $$ at $$ a = 2 $$ is: $$ P_3(x) = \frac{1}{2} \ln 2 + \frac{1}{4}(x - 2) - \frac{1}{16}(x - 2)^2 + \frac{1}{48}(x - 2)^3 $$

ind the Taylor polynomial of order 3 generated by at the point

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