How to Find the Maclaurin Series for f(x) = cos(x^3) and Determine f(6)?

To find the Maclaurin series for the function f(x) = cos(x³), we start with the known Maclaurin series expansion for cos(x), which is:

$$
cos(x) = rac{1}{0!} – rac{x^2}{2!} + rac{x^4}{4!} – rac{x^6}{6!} + rac{x^8}{8!} – ext{…}
$$

Now, substitute x³ into this series:

$$
cos(x^3) = 1 – rac{(x^3)^2}{2!} + rac{(x^3)^4}{4!} – rac{(x^3)^6}{6!} + ext{…}
$$

This simplifies to:

$$
cos(x^3) = 1 – rac{x^6}{2!} + rac{x^{12}}{4!} – rac{x^{18}}{6!} + ext{…}
$$

Now, we have the Maclaurin series for cos(x³). Next, we want to determine f(6), which is the value of the series evaluated at x = 6. However, since this series contains terms with powers of x, we need to compute the first few terms to get an approximate value.

Calculating the first few terms:

• When n=0: $1$
• When n=1: $-rac{6^6}{2!} = -rac{46656}{2} = -23328$
• When n=2: $+rac{6^{12}}{4!} = +rac{2176782336}{24} = +90782699$

Now we can add these values:

$$f(6) ≈ 1 – 23328 + 90782699 ≈ 90759372$$

Thus, using the Maclaurin series for f(x) = cos(x³), we find that:

f(6) ≈ 90759372