The problem provided asks to approximate $63^{1/3}$ using a Taylor polynomial.

Let's break it down step-by-step:

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

To approximate $63^{1/3}$, we can choose the function $f(x) = x^{1/3}$. This function gives the cube root of $x$, which directly applies to the given expression.

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

To find a basepoint $a$, we want to select a value near 63 for which we can easily compute the cube root. A good choice is $a = 64$, because $64^{1/3} = 4$. This makes calculations simpler and keeps the approximation close to the target value.

(c) Find $T_1(x)$:

The first-degree Taylor polynomial $T_1(x)$ around $x = a$ is given by: $T_1(x) = f(a) + f'(a)(x - a)$ For $f(x) = x^{1/3}$, we first compute $f'(x)$: $f'(x) = \frac{1}{3}x^{-2/3}$ Now, using $a = 64$: $f(64) = 64^{1/3} = 4$ $f'(64) = \frac{1}{3} \cdot 64^{-2/3} = \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{48}$ Thus, the first-degree Taylor polynomial is: $T_1(x) = 4 + \frac{1}{48}(x - 64)$

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

Now, substitute $x = 63$ into the Taylor polynomial $T_1(x)$: $T_1(63) = 4 + \frac{1}{48}(63 - 64) = 4 + \frac{1}{48}(-1) = 4 - \frac{1}{48} = 4 - 0.02083 = 3.97917$ Thus, the estimated value of $63^{1/3}$ is approximately $3.97917$.

Would you like more details on any of the steps or further clarifications?

Here are 5 questions that extend the discussion:

1. How can the accuracy of this approximation be improved with higher-order terms?
2. What is the general formula for the $n$-th derivative of $f(x) = x^{1/3}$?
3. Why is $a = 64$ a better choice than $a = 60$ or $a = 65$ for this approximation?
4. How would the approximation differ if you used a second-degree Taylor polynomial?
5. Can this method be used to approximate other roots, like $63^{1/5}$?

Tip: When using Taylor polynomials, selecting a basepoint $a$ close to the value you're approximating improves the accuracy of the approximation!