Find&nbsp;the&nbsp;Taylor&nbsp;polynomial&nbsp;&nbsp;T3(x)&nbsp;&nbsp;for&nbsp;the&nbsp;function&nbsp;f&nbsp;centered&nbsp;at&nbsp;the&nbsp;number&nbsp;a.&nbsp;f(x)&nbsp;=&nbsp;xe-2x,&nbsp;a&nbsp;=&nbsp;0

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Find&amp;nbsp;the&amp;nbsp;Taylor&amp;nbsp;polynomial&amp;nbsp;&amp;nbsp;T3(x)&amp;nbsp;&amp;nbsp;for&amp;nbsp;the&amp;nbsp;function&amp;nbsp;f&amp;nbsp;centered&amp;nbsp;at&amp;nbsp;the&amp;nbsp;number&amp;nbsp;a.&amp;nbsp;f(x)&amp;nbsp;=&amp;nbsp;xe-2x,&amp;nbsp;a&amp;nbsp;=&amp;nbsp;0

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To find the Taylor polynomial T3(x) for the function f(x)=xe−2x centered at a=0, we need to find the derivatives of f(x) at x=0.First, let&#039;s find the first few derivatives of f(x):f(x)=xe−2xFirst derivative:f′(x)=(1)(e−2x)+(x)(−2e−2x)=e−2x−2xe−2xSecond derivative:f″(x)=(−2)(e−2x)+(e−2x)(−2)+(−2x)(−2e−2x)=−4e−2x+4xe−2xThird derivative:f″′(x)=(4)(e−2x)+(−4)(e−2x)+(4x)(−2e−2x)+(4)(−2e−2x)=−16e−2x+12xe−2xNow, let&#039;s evaluate these derivatives at x=0:f(0)=0e0=0f′(0)=e0−2(0)e0=1f″(0)=−4e0+4(0)e0=−4f″′(0)=−16e0+12(0)e0=−16Now, we can construct the Taylor polynomial T3(x) centered at a=0 using these derivatives:T3(x)=f(0)+f′(0)(x−0)+f″(0)2!(x−0)2+f″′(0)3!(x−0)3Simplifying:T3(x)=0+1(x)+−42!(x2)+−163!(x3)T3(x)=x−2x2−83x3Therefore, the Taylor polynomial T3(x) for the function f(x)=xe−2x centered at a=0 is:T3(x)=x−2x2−83x3

Find the Taylor polynomial  T3(x)  for the function f centered at the number a. f(x) = xe-2x, a = 0

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