91Ó°ÊÓ

We have the given function $f\left(x\right)=\sqrt[3]{x}$with a derivative of order $4$at$x=1$.

The fourth taylor polynomial for $x=1$is given by,

$P_4\left(x\right)=f\left(1\right)+f^{\mathrm{′}}\left(1\right)(x−1)+\frac{f^{′â\P IJ}\left(1\right)}{2!}(x−1{)}^2+\frac{f^{′â\P IJ}\left(1\right)}{3!}(x−1{)}^3+\frac{f^{′â\P IJ′â\P IJ}\left(1\right)}{4!}(x−1{)}^4$

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 $x=1$.

The value of the function atlocalid="1649396841451" $x=1$islocalid="1649396836147" $f\left(1\right)=\sqrt[3]{1}=1$

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

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

So, at $x=1$

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

$\begin{array}{rl}{f}^{′â\P IJ}\left(x\right)& =\frac{d}{dx}\left[\frac{1}{3}{x}^{−\frac{2}{3}}\right]\\ & =\frac{1}{3}\frac{d}{dx}\left[x^{−\frac{2}{3}}\right]\\ & =\frac{1}{3}â‹\dots −\frac{2}{3}(x{)}^{−\frac{5}{3}}\\ & =−\frac{2}{9}{x}^{−\frac{5}{3}}\end{array}$

So at,$x=1$

$\begin{array}{rl}{f}^{′â\P IJ}\left(1\right)& =−\frac{2}{9}(1{)}^{−\frac{5}{3}}\\ & =−\frac{2}{9}\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(x\right)& =\frac{d}{dx}[−\frac{2}{9}{x}^{−\frac{5}{3}}]\\ & =−\frac{2}{9}\frac{d}{dx}{x}^{−\frac{5}{3}}\\ & =−\frac{2}{9}â‹\dots −\frac{5}{3}{x}^{−\frac{8}{3}}\\ & =\frac{10}{27}{x}^{−\frac{8}{3}}\end{array}$

So, at $x=1$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(1\right)& =\frac{10}{27}(1{)}^{−\frac{8}{3}}\\ & =\frac{10}{27}\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(x\right)& =\frac{d}{dx}\left[\frac{10}{27}{x}^{−\frac{8}{3}}\right]\\ & =\frac{10}{27}â‹\dots −\frac{8}{3}{x}^{−\frac{11}{3}}\\ & =−\frac{80}{81}{x}^{−\frac{11}{3}}\end{array}$

So, at $x=1$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(1\right)& =−\frac{80}{81}(1{)}^{−\frac{11}{3}}\\ & =−\frac{80}{81}\end{array}$

Hence the fourth Taylor polynomial of the function $f\left(x\right)=\sqrt[3]{x}$at $x=1$

$\begin{array}{rl}{P}_4\left(x\right)& =1+\frac{1}{3}(x−1)+\frac{−\frac{2}{9}}{2!}(x−1{)}^2+\frac{\frac{10}{27}}{3!}(x−1{)}^3+\frac{−\frac{80}{81}}{4!}(x−1{)}^4\\ & =1+\frac{1}{3}(x−1)−\frac{1}{9}(x−1{)}^2+\frac{5}{81}(x−1{)}^3−\frac{10}{243}(x−1{)}^4\end{array}$