Q. 45 - Calculus

Question

In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$ for the specified function and the given value of $x_0$.

$45. \ln \left(x\right),3$

Step-by-Step Solution

Answer

The fourth Taylor polynomial of the function$f\left(x\right)=\ln \left(x\right)$ at $x=3$ is, $P_4\left(x\right)=\ln 3+\frac{1}{3}(x-3)-\frac{1}{18}(x-3{)}^2+\frac{1}{81}(x-3{)}^3-\frac{1}{324}(x-3{)}^4$

Step 1. Given data

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

Step 2. The fourth taylor polynomial

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$f\left(3\right)=\ln \left(3\right)$

Step 3. Find f ' ( x )

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

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

So, at $x=3$

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

Step 4. Find f ' ' ( x )

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

So, at $x=3$

$\begin{array}{rl}{f}^{\prime \prime }\left(3\right)& =-\frac{1}{3^2}\\ & =-\frac{1}{9}\end{array}$

Step 5. Find f ' ' ' ( x )

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

So, at $x=3$

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

Step 6. Find f ' ' ' ' ( x )

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

So, at $x=3$

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

Step 7. The fourth Taylor polynomial of the function

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

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