Question

b8-taylor and maclaurin series: problem 1
(2 points)
write out the first four terms of the maclaurin series of ( f(x) ) if
( f(0)=3, quad f^{prime}(0)=3, quad f^{prime prime}(0)=-14, quad f^{prime prime prime}(0)=-13 )
( f(x)= )+
preview my answers submit answers
you have attempted this problem 0 times.
you have 15 attempts remaining.
page generated october 21, 2025, 10:29:08 pm edt
webwork ( subset 1996 - 2025 ) | theme: math4 | ww_version: 2.20 | pg_version 2.20
the webwork project

Explanation:

Step1: Recall Maclaurin series formula

The Maclaurin series of a function \(f(x)\) is given by \(f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{\prime\prime\prime}(0)}{3!}x^{3}+\cdots\)

Step2: Substitute the given values

We are given \(f(0) = 3\), \(f^{\prime}(0)=3\), \(f^{\prime\prime}(0)=- 14\), \(f^{\prime\prime\prime}(0)=-13\)

For \(n = 0\): \(\frac{f^{(0)}(0)}{0!}x^{0}=f(0)=3\) (since \(0!=1\) and \(x^{0} = 1\))

For \(n = 1\): \(\frac{f^{\prime}(0)}{1!}x^{1}=3x\) (since \(1!=1\))

For \(n = 2\): \(\frac{f^{\prime\prime}(0)}{2!}x^{2}=\frac{-14}{2}x^{2}=-7x^{2}\) (since \(2!=2\times1 = 2\))

For \(n = 3\): \(\frac{f^{\prime\prime\prime}(0)}{3!}x^{3}=\frac{-13}{6}x^{3}\) (since \(3!=3\times2\times1=6\))

Answer:

\(3 + 3x-7x^{2}-\frac{13}{6}x^{3}\)