work problem 1. find the problem with the ans...

# Explanation: ## Step1: Recall Taylor - polynomial formula The $n$-th degree Taylor polynomial of a function $f(x)$ about $x = a$ is given by $P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x - a)^n$. ## Step2: Identify the fourth - term of the third - degree Taylor polynomial For the third - degree Taylor polynomial $P_3(x)$ of a function $f(x)$ about $x = 3$, the formula is $P_3(x)=f(3)+f^{\prime}(3)(x - 3)+\frac{f^{\prime\prime}(3)}{2!}(x - 3)^2+\frac{f^{(3)}(3)}{3!}(x - 3)^3$. The fourth term is $\frac{f^{(3)}(3)}{3!}(x - 3)^3$. ## Step3: Substitute the given values We are given that $f(3)=2,f^{\prime}(3)=-5,f^{\prime\prime}(3)=9,f^{(3)}(3)=4$. Substituting $f^{(3)}(3) = 4$ into the fourth - term formula $\frac{f^{(3)}(3)}{3!}(x - 3)^3$, we have $\frac{4}{3!}(x - 3)^3=\frac{4}{6}(x - 3)^3=\frac{2}{3}(x - 3)^3$. The coefficient of $(x - 3)^3$ is $\frac{4}{6}=\frac{2}{3}$, not $\frac{1}{150}$. There seems to be an error in the provided work. But if we just focus on finding the fourth term of the third - degree Taylor polynomial using the correct formula: The fourth term of the third - degree Taylor polynomial for $f$ about $x = 3$ is $\frac{f^{(3)}(3)}{3!}(x - 3)^3$. Since $f^{(3)}(3)=4$ and $3!=6$, the fourth term is $\frac{4}{6}(x - 3)^3=\frac{2}{3}(x - 3)^3$. # Answer: $\frac{2}{3}(x - 3)^3$