Question

question 4 (1 point)
which of the following is the degree 3 taylor polynomial of $f(x)=ln(x)$ at $x = 7$?
$ln(7)+\frac{1}{7}x-\frac{1}{49}x^{2}+\frac{2}{343}x^{3}$
$ln(7)+\frac{1}{7}(x - 7)-\frac{1}{98}(x - 7)^{2}+\frac{1}{1029}(x - 7)^{3}$
$ln(7)+\frac{1}{7}(x - 7)-\frac{1}{49}(x - 7)^{2}+\frac{2}{343}(x - 7)^{3}$
none of these options are correct.
$ln(7)+\frac{1}{7}x+\frac{1}{98}x^{2}-\frac{1}{1029}x^{3}$
view hint for question 4

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ about $x = a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\frac{f^{(3)}(a)}{3!}(x - a)^3+\cdots$, where $f^{(n)}(a)$ is the $n$ - th derivative of $f(x)$ evaluated at $x = a$.

Step2: Find the derivatives of $f(x)=\ln(x)$

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

The degree - 3 Taylor polynomial $P_3(x)$ of $f(x)=\ln(x)$ about $x = 7$ is $P_3(x)=f(7)+f^{\prime}(7)(x - 7)+\frac{f^{\prime\prime}(7)}{2}(x - 7)^2+\frac{f^{(3)}(7)}{6}(x - 7)^3$.
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:

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