question 4 (1 point)\nwhich of the following is the degree 3 taylor polynomial of $f(x)=ln(x)$ at $x =…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ about $x = a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}+\cdots$. For $f(x)=\ln(x)$ and $a = 7$.

Step2: Calculate derivatives

First - derivative: $f^{\prime}(x)=\frac{1}{x}$, so $f^{\prime}(7)=\frac{1}{7}$. Second - derivative: $f^{\prime\prime}(x)=-\frac{1}{x^{2}}$, so $f^{\prime\prime}(7)=-\frac{1}{49}$. Third - derivative: $f^{(3)}(x)=\frac{2}{x^{3}}$, so $f^{(3)}(7)=\frac{2}{343}$. And $f(7)=\ln(7)$.

Step3: Construct the degree - 3 Taylor polynomial

$P_3(x)=f(7)+f^{\prime}(7)(x - 7)+\frac{f^{\prime\prime}(7)}{2}(x - 7)^{2}+\frac{f^{(3)}(7)}{6}(x - 7)^{3}$. Substitute the values: $P_3(x)=\ln(7)+\frac{1}{7}(x - 7)-\frac{1}{98}(x - 7)^{2}+\frac{1}{1029}(x - 7)^{3}$.

Answer:

$\ln(7)+\frac{1}{7}(x - 7)-\frac{1}{98}(x - 7)^{2}+\frac{1}{1029}(x - 7)^{3}$