Austin Lohr

2022-10-14

star233

To find the Taylor polynomial ${T}_{3}\left(x\right)$ for the function $f\left(x\right)=x{e}^{-2x}$ centered at $a=0$, we need to find the derivatives of $f\left(x\right)$ at $x=0$.
First, let's find the first few derivatives of $f\left(x\right)$:
$f\left(x\right)=x{e}^{-2x}$
First derivative:
${f}^{\prime }\left(x\right)=\left(1\right)\left({e}^{-2x}\right)+\left(x\right)\left(-2{e}^{-2x}\right)={e}^{-2x}-2x{e}^{-2x}$
Second derivative:
${f}^{″}\left(x\right)=\left(-2\right)\left({e}^{-2x}\right)+\left({e}^{-2x}\right)\left(-2\right)+\left(-2x\right)\left(-2{e}^{-2x}\right)=-4{e}^{-2x}+4x{e}^{-2x}$
Third derivative:
${{f}^{″}}^{\prime }\left(x\right)=\left(4\right)\left({e}^{-2x}\right)+\left(-4\right)\left({e}^{-2x}\right)+\left(4x\right)\left(-2{e}^{-2x}\right)+\left(4\right)\left(-2{e}^{-2x}\right)=-16{e}^{-2x}+12x{e}^{-2x}$
Now, let's evaluate these derivatives at $x=0$:
$f\left(0\right)=0{e}^{0}=0$
${f}^{\prime }\left(0\right)={e}^{0}-2\left(0\right){e}^{0}=1$
${f}^{″}\left(0\right)=-4{e}^{0}+4\left(0\right){e}^{0}=-4$
${{f}^{″}}^{\prime }\left(0\right)=-16{e}^{0}+12\left(0\right){e}^{0}=-16$
Now, we can construct the Taylor polynomial ${T}_{3}\left(x\right)$ centered at $a=0$ using these derivatives:
${T}_{3}\left(x\right)=f\left(0\right)+{f}^{\prime }\left(0\right)\left(x-0\right)+\frac{{f}^{″}\left(0\right)}{2!}\left(x-0{\right)}^{2}+\frac{{{f}^{″}}^{\prime }\left(0\right)}{3!}\left(x-0{\right)}^{3}$
Simplifying:
${T}_{3}\left(x\right)=0+1\left(x\right)+\frac{-4}{2!}\left({x}^{2}\right)+\frac{-16}{3!}\left({x}^{3}\right)$
${T}_{3}\left(x\right)=x-2{x}^{2}-\frac{8}{3}{x}^{3}$
Therefore, the Taylor polynomial ${T}_{3}\left(x\right)$ for the function $f\left(x\right)=x{e}^{-2x}$ centered at $a=0$ is:
${T}_{3}\left(x\right)=x-2{x}^{2}-\frac{8}{3}{x}^{3}$

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