Problem 1: Find the degree 3 Taylor Polynomial for $f(x) = \sqrt{x}$ centered at $a = 9$ and use it to approximate $\sqrt{12}$.

The Taylor polynomial of degree 3 for a function $f(x)$ centered at $a$ is given by:

$T_3(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f^{(3)}(a)}{3!}(x - a)^3$

Let's apply this to the function $f(x) = \sqrt{x}$, with $a = 9$.

Step 1: Compute derivatives of $f(x) = \sqrt{x}$

1. $f(x) = x^{1/2}$
2. $f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$
3. $f''(x) = -\frac{1}{4} x^{-3/2} = -\frac{1}{4x^{3/2}}$
4. $f^{(3)}(x) = \frac{3}{8} x^{-5/2} = \frac{3}{8x^{5/2}}$

Step 2: Evaluate the derivatives at $x = 9$

1. $f(9) = \sqrt{9} = 3$
2. $f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$
3. $f''(9) = -\frac{1}{4(9)^{3/2}} = -\frac{1}{4(27)} = -\frac{1}{108}$
4. $f^{(3)}(9) = \frac{3}{8(9)^{5/2}} = \frac{3}{8(243)} = \frac{1}{648}$

Step 3: Write the Taylor polynomial

Now substitute the values into the Taylor polynomial formula:

$T_3(x) = 3 + \frac{1}{6}(x - 9) - \frac{1}{108}(x - 9)^2 + \frac{1}{648}(x - 9)^3$

Step 4: Use the Taylor polynomial to approximate $\sqrt{12}$

Now, to approximate $\sqrt{12}$, substitute $x = 12$ into the Taylor polynomial:

$T_3(12) = 3 + \frac{1}{6}(12 - 9) - \frac{1}{108}(12 - 9)^2 + \frac{1}{648}(12 - 9)^3$ $T_3(12) = 3 + \frac{1}{6}(3) - \frac{1}{108}(3)^2 + \frac{1}{648}(3)^3$ $T_3(12) = 3 + \frac{1}{6}(3) - \frac{1}{108}(9) + \frac{1}{648}(27)$ $T_3(12) = 3 + \frac{3}{6} - \frac{9}{108} + \frac{27}{648}$ $T_3(12) = 3 + 0.5 - 0.0833 + 0.04167$ $T_3(12) \approx 3.4583$

Thus, the approximation for $\sqrt{12}$ is approximately $3.4583$.

Problem 2: Use Taylor's Remainder Theorem to find an upper bound for the approximation.

Taylor's Remainder Theorem states that the remainder term $R_3(x)$ for a degree 3 Taylor polynomial is given by:

$R_3(x) = \frac{f^{(4)}(c)}{4!} (x - a)^4$

for some $c$ between $a$ and $x$. Let's find an upper bound for the remainder.

Step 1: Compute the fourth derivative of $f(x) = \sqrt{x}$

$f^{(4)}(x) = -\frac{15}{16} x^{-7/2}$

Step 2: Evaluate $f^{(4)}(c)$

We need an upper bound on $f^{(4)}(c)$ for $c \in [9, 12]$. Since $f^{(4)}(x) = -\frac{15}{16} x^{-7/2}$, the value of $|f^{(4)}(x)|$ is maximized when $x = 9$. So, let's evaluate it at $x = 9$:

$|f^{(4)}(9)| = \frac{15}{16(9)^{7/2}} = \frac{15}{16(729)} = \frac{15}{11664}$

Step 3: Compute the remainder

The remainder $R_3(12)$ is:

$R_3(12) = \frac{f^{(4)}(c)}{4!} (12 - 9)^4 = \frac{15}{11664 \times 24} \times 3^4$ $R_3(12) = \frac{15}{279936} \times 81 = \frac{1215}{279936} \approx 0.00434$

Thus, the upper bound for the remainder is approximately $0.00434$.

Do you need more details on any step? Here are some related questions for you to explore:

1. How can you use Taylor polynomials to estimate functions at different points?
2. Why does the degree of the Taylor polynomial affect the accuracy of the approximation?
3. What are other methods besides Taylor polynomials to approximate $\sqrt{12}$?
4. How does the Taylor Remainder Theorem ensure accuracy within certain bounds?
5. How would the approximation change if a higher-degree polynomial were used?

Tip: To improve accuracy in using Taylor polynomials, increasing the degree of the polynomial will typically reduce the remainder.