# Find the Taylor polynomial of degree n=4 for each function expanded about the given value of x_0 f(x)=x^5+4x^2+3x+1, x_0 =0

Find the Taylor polynomial of degree n=4 for each function expanded about the given value of ${x}_{0}$.
$f\left(x\right)={x}^{5}+4{x}^{2}+3x+1,{x}_{0}=0$
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edgarovhg
f(0) = 1
${f}^{\prime }\left(x\right)=5{x}^{4}+8x+3$
f'(0) = 3
${f}^{″}\left(x\right)=20{x}^{3}+8$
f'''(0) = 8
${f}^{3}\left(x\right)=60{x}^{2}$
${f}^{3}\left(0\right)=0$
${f}^{4}\left(x\right)=120x,{f}^{4}\left(0\right)=0$
${f}^{5}\left(x\right)=120,{f}^{5}\left(0\right)=0$
so we have:
$f\left(x\right)=1+\frac{3}{1!}x+\frac{8}{2!}{x}^{2}+\frac{0}{3!}{x}^{3}+\frac{0}{4!}{x}^{0}+\frac{120}{5!}{x}^{5}=1+3x+4{x}^{2}+{x}^{5}$
since we only consider n=4, we have:
$f\left(x\right)=1+3x+4{x}^{2}$