Find the Taylor polynomial of degree n=4 for each function expanded about the given value of ${x}_{0}$.

$f(x)={x}^{5}+4{x}^{2}+3x+1,{x}_{0}=0$

$f(x)={x}^{5}+4{x}^{2}+3x+1,{x}_{0}=0$

Greyson Landry
2022-07-30
Answered

Find the Taylor polynomial of degree n=4 for each function expanded about the given value of ${x}_{0}$.

$f(x)={x}^{5}+4{x}^{2}+3x+1,{x}_{0}=0$

$f(x)={x}^{5}+4{x}^{2}+3x+1,{x}_{0}=0$

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edgarovhg

Answered 2022-07-31
Author has **12** answers

f(0) = 1

${f}^{\prime}(x)=5{x}^{4}+8x+3$

f'(0) = 3

${f}^{\u2033}(x)=20{x}^{3}+8$

f'''(0) = 8

${f}^{3}(x)=60{x}^{2}$

${f}^{3}(0)=0$

${f}^{4}(x)=120x,{f}^{4}(0)=0$

${f}^{5}(x)=120,{f}^{5}(0)=0$

so we have:

$f(x)=1+\frac{3}{1!}x+\frac{8}{2!}{x}^{2}+\frac{0}{3!}{x}^{3}+\frac{0}{4!}{x}^{0}+\frac{120}{5!}{x}^{5}=1+3x+4{x}^{2}+{x}^{5}$

since we only consider n=4, we have:

$f(x)=1+3x+4{x}^{2}$

${f}^{\prime}(x)=5{x}^{4}+8x+3$

f'(0) = 3

${f}^{\u2033}(x)=20{x}^{3}+8$

f'''(0) = 8

${f}^{3}(x)=60{x}^{2}$

${f}^{3}(0)=0$

${f}^{4}(x)=120x,{f}^{4}(0)=0$

${f}^{5}(x)=120,{f}^{5}(0)=0$

so we have:

$f(x)=1+\frac{3}{1!}x+\frac{8}{2!}{x}^{2}+\frac{0}{3!}{x}^{3}+\frac{0}{4!}{x}^{0}+\frac{120}{5!}{x}^{5}=1+3x+4{x}^{2}+{x}^{5}$

since we only consider n=4, we have:

$f(x)=1+3x+4{x}^{2}$

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