91Ó°ÊÓ

We have the given function $f\left(x\right)=\ln \left(x\right)$with a derivative of order$4$at$x=3$.

The fourth taylor polynomial for $x=3$is given by,

$P_4\left(x\right)=f\left(3\right)+f^{\text{'}}\left(3\right)(x-3)+\frac{f^{\text{'}\text{'}}\left(3\right)}{2!}(x-3{)}^2+\frac{f^{\text{'}\text{'}}\left(3\right)}{3!}(x-3{)}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(3\right)}{4!}(x-3{)}^4$

Therefore, we have to find the value of the function along with$f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and$f\text{'}\text{'}\text{'}\text{'}\left(x\right)$at $x=3$

The value of the function at $x=3$is$f\left(3\right)=\ln \left(3\right)$

The derivatives of the function,$f\left(x\right)=\ln \left(x\right)$

$\begin{array}{rl}{f}^{\mathrm{′}}\left(x\right)& =\frac{d}{dx}\left[\ln \right(x\left)\right]\\ & =\frac{1}{x}\end{array}$

So, at $x=3$

$f^{\mathrm{′}}\left(3\right)=\frac{1}{3}$

$\begin{array}{rl}{f}^{′â\P IJ}\left(x\right)& =\frac{d}{dx}\left[\frac{1}{x}\right]\\ & =−\frac{1}{x^2}\end{array}$

So, at $x=3$

$\begin{array}{rl}{f}^{′â\P IJ}\left(3\right)& =−\frac{1}{3^2}\\ & =−\frac{1}{9}\end{array}$

$\begin{array}{rl}f\text{'}\text{'}\text{'}\left(x\right)& =\frac{d}{dx}[−\frac{1}{x^2}]\\ & =−\frac{d}{dx}[x{]}^{−2}\\ & =−\frac{−2}{x^3}\\ & =\frac{2}{x^3}\end{array}$

So, at$x=3$

localid="1649392927957" $\begin{array}{rl}f\text{'}\text{'}\text{'}\left(3\right)& =\frac{2}{3^3}\\ & =\frac{2}{27}\end{array}$

$\begin{array}{rl}f\text{'}\text{'}\text{'}\text{'}\left(x\right)& =\frac{d}{dx}\left[\frac{2}{x^3}\right]\\ & =2\frac{d}{dx}{x}^{−3}\\ & =2\frac{−3}{x^4}\\ & =−\frac{6}{x^4}\end{array}$

So, at $x=3$

$\begin{array}{rl}f\text{'}\text{'}\text{'}\text{'}\left(3\right)& =−\frac{6}{3^4}\\ & =−\frac{2}{27}\end{array}$

Hence the fourth Taylor polynomial of the function $f\left(x\right)=\ln \left(x\right)$at$x=3$is,

role="math" localid="1649392862804" $\begin{array}{rl}{P}_4\left(x\right)& =\ln 3+\frac{1}{3}(x−3)+\frac{−\frac{1}{2}}{2!}(x−3{)}^2+\frac{\frac{2}{27}}{3!}(x−3{)}^3+\frac{−\frac{2}{27}}{4!}(x−3{)}^4\\ & =\ln 3+\frac{1}{3}(x−3)−\frac{1}{18}(x−3{)}^2+\frac{1}{81}(x−3{)}^3−\frac{1}{324}(x−3{)}^4\end{array}$