To find the Taylor series for $f(x) = \sqrt{x}$ centered at $x = 1$, we use the formula for a Taylor series expansion:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x - 1)^n$

We need to calculate the first few derivatives of $f(x) = \sqrt{x}$ at $x = 1$ and evaluate them at $x = 1$.

1. First, the function itself: $f(x) = \sqrt{x} = x^{1/2}$ $f(1) = \sqrt{1} = 1$

2. First derivative: $f'(x) = \frac{1}{2} x^{-1/2}$ $f'(1) = \frac{1}{2} \cdot 1^{-1/2} = \frac{1}{2}$

3. Second derivative: $f''(x) = \frac{-1}{4} x^{-3/2}$ $f''(1) = \frac{-1}{4} \cdot 1^{-3/2} = \frac{-1}{4}$

4. Third derivative: $f^{(3)}(x) = \frac{3}{8} x^{-5/2}$ $f^{(3)}(1) = \frac{3}{8} \cdot 1^{-5/2} = \frac{3}{8}$

5. Fourth derivative: $f^{(4)}(x) = \frac{-15}{16} x^{-7/2}$ $f^{(4)}(1) = \frac{-15}{16} \cdot 1^{-7/2} = \frac{-15}{16}$

Now, we can plug these values into the Taylor series formula:

$f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f^{(3)}(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4 + \cdots$

Substituting the calculated values:

$\sqrt{x} = 1 + \frac{1}{2}(x - 1) + \frac{-1}{4} \frac{(x - 1)^2}{2!} + \frac{3}{8} \frac{(x - 1)^3}{3!} + \frac{-15}{16} \frac{(x - 1)^4}{4!} + \cdots$

Simplifying:

$\sqrt{x} = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2 + \frac{1}{16}(x - 1)^3 - \frac{5}{128}(x - 1)^4 + \cdots$

Thus, the first five terms of the Taylor series for $\sqrt{x}$ centered at $x = 1$ are:

$1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2 + \frac{1}{16}(x - 1)^3 - \frac{5}{128}(x - 1)^4$

Do you want more details or have any questions?

Here are some related questions to expand your understanding:

1. How do you find higher-order terms in a Taylor series?
2. What conditions ensure that a Taylor series converges to the original function?
3. How is a Taylor series related to a Maclaurin series?
4. Can you compute the Taylor series for $\sqrt{x}$ around a different center, say $x = 4$?
5. How can you use the Taylor series to approximate $\sqrt{1.1}$?

Tip: A Taylor series approximates a function around a specific point. The more terms you use, the more accurate the approximation near that point.