Let's address each part of the problem using a Taylor polynomial approximation for $63^{1/3}$.

(a) Determine an appropriate power function $f(x)$.

Since we want to approximate $63^{1/3}$, a suitable choice for $f(x)$ would be: $f(x) = x^{1/3}$

(b) What would be an appropriate basepoint $a$?

The basepoint $a$ should be a value close to 63 that makes the calculations simpler and for which we can easily find the cube root. A good choice here is $a = 64$, since $64^{1/3} = 4$.

(c) Find $T_1(x)$.

The Taylor polynomial of degree 1 (linear approximation) centered at $a = 64$ for the function $f(x) = x^{1/3}$ is given by: $T_1(x) = f(a) + f'(a)(x - a)$

1. Calculate $f(a)$: $f(64) = 64^{1/3} = 4$

2. Calculate $f'(x)$: The derivative of $f(x) = x^{1/3}$ is: $f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3 \cdot x^{2/3}}$

3. Evaluate $f'(64)$: $f'(64) = \frac{1}{3 \cdot 64^{2/3}} = \frac{1}{3 \cdot 16} = \frac{1}{48}$

4. Write $T_1(x)$: $T_1(x) = 4 + \frac{1}{48}(x - 64)$

(d) Use $T_1(x)$ to estimate the value of $63^{1/3}$.

Substitute $x = 63$ into $T_1(x)$: $T_1(63) = 4 + \frac{1}{48}(63 - 64) = 4 - \frac{1}{48} = 4 - 0.02083 \approx 3.9792$

So, the estimate for $63^{1/3}$ using the Taylor polynomial $T_1(x)$ is approximately $3.9792$.

Would you like more details on any of the steps or have any questions?

Related Questions:

1. How would the approximation improve if we used a second-degree Taylor polynomial $T_2(x)$?
2. What is the actual value of $63^{1/3}$, and how close is the Taylor polynomial approximation?
3. How do higher-order derivatives affect the accuracy of a Taylor polynomial?
4. Can you use a different basepoint $a$ for the Taylor expansion, like $a = 60$, and how would that change the result?
5. What other methods can be used to approximate $63^{1/3}$ besides Taylor series?

Tip:

For better approximations, using a basepoint $a$ that is very close to the desired value $x$ often results in a more accurate Taylor polynomial approximation, especially when using only a few terms.