answer: 3 #____ - find the coefficient of the...

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