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Gala2k [10]
3 years ago
11

Find the taylor polynomial t3(x) for the function f centered at the number

Mathematics
1 answer:
Vlad [161]3 years ago
5 0
e^{-3x}=\displaystyle\sum_{n=0}^\infty\frac{(-3x)^n}{n!}=1-3x+9x^2+\cdots
\sin2x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n+1}}{(2n+1)!}=2x-\dfrac{4x^3}3+\cdots

e^{-3x}\sin2x=\left(1-3x+9x^2+\cdots\right)\left(2x-\dfrac{4x^3}3+\cdots\right)
\approx T_3(x)=(1-3x+9x^2)\left(2x-\dfrac{4x^3}3\right)
T_3(x)=2x-6x^2+\left(18-\dfrac43\right)x^3
T_3(x)=2x-6x^2+\dfrac{50}3x^3
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Step-by-step explanation:

1.

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(-1,7) =(x_1,y_1)\\\\5y = 4x+5\\\\Write\:in\:y=mx+c\:form\\\frac{5y}{5} = \frac{4}{5} x + \frac{5}{5} \\y = \frac{4}{5} x +1\\m = \frac{4}{5} \\\\y -y_1=m(x-x_1)\\y-7=\frac{4}{5} (x -(-1)\\y -7 = 4/5(x+1)\\y -7 = 4/5x +4/5\\\\y = \frac{4}{5} x+\frac{4}{5} +7\\\\y = \frac{4}{5} x +39/5

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To solve an equation or an inequality, you do the opposite operation in the opposite order.

Since the second operation was add 2/5, to solve, you start by doing the opposite of the second operation. That means you subtract 2/5 from both sides.

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