With Taylor/Maclaurin polynomials

Use the Remainder Estimation Theorem to find an interval containing x = 0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.

f(x) = sin x
p(x) = x - 1/6 * x^3

The book answer is (-.569, +.569)
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My work so far

For three decimal-place accuracy
R(x) <= 5 * 10^-4

For MacLaurin of x - 1/6 * x^3, n must equal 4 so
R(x) <= M/24 * x^4

f to the 4th derivative over interval <= M
sin x over interval <= M

R(x) <= sin x / 24 * x^4
5*10^-4 >= sin x / 24 * x^4

Is this right so far? I can use a computer to find several solutions to the above equation but they don't match the book's answer.

Your approach so far is correct, but it seems there might be a misunderstanding in the application of the Remainder Estimation Theorem. Let's go through the correct steps to find the interval containing x = 0 over which f(x) can be approximated by p(x) with three decimal-place accuracy throughout the interval.

1. The Maclaurin polynomial for f(x) = sin x is given by p(x) = x - (1/6)x^3.

2. To find an interval of accuracy, we need to find the remainder term R(x) for this polynomial approximation.

3. The Remainder Estimation Theorem states that for a function f(x) and its related Maclaurin polynomial approximation p(x), the remainder term R(x) is given by:

R(x) = f^(n+1)(c) / (n+1)! * (x-a)^(n+1)

where f^(n+1)(c) is the (n+1)th derivative of f(x) evaluated at some point c between a and x.

4. In this case, we have p(x) = x - (1/6)x^3, which means n = 3 since the highest power of x in the polynomial is 3.

5. We need to find an upper bound for the remainder term R(x), which will give us an interval of accuracy.

6. The fourth derivative of f(x) = sin x is f^(4)(x) = sin x, which implies that M = 1.

7. Therefore, the remainder term R(x) for the given polynomial approximation is:

R(x) = sin(c) / 4! * x^4

8. We want R(x) to be less than or equal to 5 * 10^(-4) for three decimal-place accuracy. So we have:

sin(c) / 24 * x^4 <= 5 * 10^(-4)

9. To find the interval containing x = 0, we need to determine the maximum value of sin(c) and x^4 over the entire interval.

10. Since the sine function has a maximum value of 1, we can simplify the inequality to:

1 / 24 * x^4 <= 5 * 10^(-4)

11. Solving this inequality gives us the interval:

-0.569 <= x <= 0.569

So, the correct interval containing x = 0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval is (-0.569, 0.569), which matches the book's answer.