b8-taylor and maclaurin series: problem 1\n(2 points)\nwrite out the first four terms of the maclaurin…

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

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

Answer

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})