answer: 3 #____ - find the coefficient of the third - degree term in the taylor series for f(x)=e^3x…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n$. The $n$-th term is $T_n=\frac{f^{(n)}(a)}{n!}(x - a)^n$. We want the third - degree term, so $n = 3$ and $a = 2$.

Step2: Find the third - derivative of $y = e^{3x}$

The first derivative of $y=e^{3x}$ using the chain - rule: if $y = e^{u}$ and $u = 3x$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}=3e^{3x}$. The second derivative $y^{\prime\prime}=\frac{d}{dx}(3e^{3x}) = 9e^{3x}$. The third derivative $y^{(3)}=\frac{d}{dx}(9e^{3x})=27e^{3x}$.

Step3: Evaluate the third - derivative at $x = 2$

Substitute $x = 2$ into $y^{(3)}(x)$. We get $y^{(3)}(2)=27e^{3\times2}=27e^{6}$.

Step4: Calculate the coefficient of the third - degree term

The coefficient of the third - degree term ($n = 3$) in the Taylor series is $\frac{f^{(3)}(2)}{3!}$. Since $f^{(3)}(2)=27e^{6}$ and $3!=6$, the coefficient is $\frac{27e^{6}}{6}=\frac{9e^{6}}{2}$.

Answer:

$\frac{9e^{6}}{2}$