Question

In Exercises $41-48$in Section $8.2$, you were asked to find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and

the given value of $x_0$. In Exercises $37-44$give Lagrange’s form for the remainder $R_4\left(x\right)$.

$\sqrt[3]{x},1$

Short answer

The required Lagrange's form is$R_4\left(x\right)=\frac{11c^{-\frac{14}{3}}}{2673}(x-1{)}^5$

Step-by-step solution

Given Information

The given function is $f\left(x\right)=\sqrt[3]{x}$

Finding the derivatives of the given function

The derivatives of the function $f\left(x\right)=\sqrt[3]{x}$are

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

Also,

$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\frac{1}{3}{x}^{-\frac{2}{3}}\right]\\ & =& \frac{1}{3}\frac{d}{dx}\left[x^{-\frac{2}{3}}\right]\\ & =& \frac{1}{3}·-\frac{2}{3}(x{)}^{-\frac{5}{3}}\\ & =& -\frac{2}{9}{x}^{-\frac{3}{3}}\end{array}$

Again,

$$\begin{gathered} f^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[-\frac{2}{9}{x}^{\frac{5}{3}}\right] \\ =-\frac{2}{9}\frac{d}{dx}{x}^{\frac{5}{3}} \\ =-\frac{2}{9}\times -\frac{5}{3}{x}^{-\frac{5}{3}} \\ =\frac{10}{27}{x}^{-\frac{5}{3}} \end{gathered}$$

Also,

$$\begin{gathered} f^{\text{'}\text{'}\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[\frac{10}{27}{x}^{-\frac{8}{3}}\right] \\ =\frac{10}{27}\times -\frac{8}{3}{x}^{\frac{11}{3}} \\ =-\frac{80}{81}{x}^{\frac{11}{3}} \end{gathered}$$

Finally,

$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[-\frac{80}{81}{x}^{-\frac{11}{3}}\right]\\ & =& -\frac{80}{81}\frac{d}{dx}\left[x^{-\frac{11}{3}}\right]\\ & =& -\frac{80}{81}\times -\frac{11}{3}{x}^{-\frac{14}{3}}\\ & =& \frac{88}{243}{x}^{-\frac{14}{3}}\end{array}$

Determine the Lagrange’s form for the remainder

Now, by the Lagrange's form for the remainder, if $f$is a function that can be differentiated $n+1$times in some open interval $I$containing the point $x_0$and $R_n\left(x\right)$be the $n$th remainder for $f$at $x=x_0$. Then there exists at least one $c$between $x_0$and $x$such that

$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Since $f^{\left(5\right)}\left(x\right)=\frac{88}{243}{x}^{-\frac{14}{3}}$and $x_0=1$then

$R_4\left(x\right)=\frac{\frac{88}{243}{c}^{-\frac{14}{3}}}{5!}(x-1{)}^5$

Final derivative

The Final remainder of derivative is${\mathrm{R}}_4\left(\mathrm{x}\right)=\frac{11{\mathrm{c}}^{-\frac{14}{3}}}{2673}(\mathrm{x}-1{)}^5$