Question

a graphing calculator is recommended.
find the taylor polynomial ( t_{3}(x) ) for the function ( f ) centered at the number ( a ).
( f(x)=e^{-3 x} sin (2 x), quad a = 0 )
( t_{3}(x)= )
graph ( f ) and ( t_{3} ) on the same screen.

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial \(T_{n}(x)\) of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by \(T_{n}(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}\). Since \(a = 0\), we have \(T_{n}(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}\), and for \(n = 3\), \(T_{3}(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{\prime\prime\prime}(0)}{3!}x^{3}\)

Step2: Find \(f(0)\)

Given \(f(x)=e^{-3x}\sin(2x)\), then \(f(0)=e^{0}\sin(0)=0\)

Step3: Find \(f^{\prime}(x)\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\)

Let \(u = e^{-3x}\) and \(v=\sin(2x)\). Then \(u^\prime=- 3e^{-3x}\) and \(v^\prime = 2\cos(2x)\)
\(f^{\prime}(x)=-3e^{-3x}\sin(2x)+2e^{-3x}\cos(2x)\)
\(f^{\prime}(0)=-3e^{0}\sin(0)+2e^{0}\cos(0)=2\)

Step4: Find \(f^{\prime\prime}(x)\)

\(f^{\prime\prime}(x)=(-3)(-3e^{-3x})\sin(2x)-3e^{-3x}(2\cos(2x))+2(-3e^{-3x})\cos(2x)+2e^{-3x}(-2\sin(2x))\)
\(f^{\prime\prime}(x)=9e^{-3x}\sin(2x)-6e^{-3x}\cos(2x)-6e^{-3x}\cos(2x)-4e^{-3x}\sin(2x)\)
\(f^{\prime\prime}(x)=(9 - 4)e^{-3x}\sin(2x)+(-6-6)e^{-3x}\cos(2x)=5e^{-3x}\sin(2x)-12e^{-3x}\cos(2x)\)
\(f^{\prime\prime}(0)=5e^{0}\sin(0)-12e^{0}\cos(0)=-12\)

Step5: Find \(f^{\prime\prime\prime}(x)\)

\(f^{\prime\prime\prime}(x)=5(-3e^{-3x})\sin(2x)+5e^{-3x}(2\cos(2x))-12(-3e^{-3x})\cos(2x)-12e^{-3x}(-2\sin(2x))\)
\(f^{\prime\prime\prime}(x)=-15e^{-3x}\sin(2x)+10e^{-3x}\cos(2x)+36e^{-3x}\cos(2x)+24e^{-3x}\sin(2x)\)
\(f^{\prime\prime\prime}(x)=(24 - 15)e^{-3x}\sin(2x)+(10 + 36)e^{-3x}\cos(2x)=9e^{-3x}\sin(2x)+46e^{-3x}\cos(2x)\)
\(f^{\prime\prime\prime}(0)=9e^{0}\sin(0)+46e^{0}\cos(0)=46\)

Step6: Construct \(T_{3}(x)\)

\(T_{3}(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2}x^{2}+\frac{f^{\prime\prime\prime}(0)}{6}x^{3}\)
Substitute \(f(0) = 0\), \(f^{\prime}(0)=2\), \(f^{\prime\prime}(0)=-12\), \(f^{\prime\prime\prime}(0)=46\)
\(T_{3}(x)=0 + 2x+\frac{-12}{2}x^{2}+\frac{46}{6}x^{3}=2x-6x^{2}+\frac{23}{3}x^{3}\)

Answer:

\(T_{3}(x)=2x - 6x^{2}+\frac{23}{3}x^{3}\)