Find the few terms of the Taylor series for the following functions about the given points.

$\sqrt[\mathbf{3}]{\mathbf{x}}\mathbf{,}\mathit{a}\mathbf{=}\mathbf{8}$

The function is $\sqrt[3]{x}\text{and}a=8$

The Taylor series ofaboutis expressed as follows:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{1}\mathbf{!}}\mathbf{(}\mathit{x}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{2}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{3}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}$

Substitute $8$for $x$in the equation $f^{\text{'}\text{'}}\left(x\right)=-\frac{2}{9}{x}^{-5/3}$ as follows:

$\begin{array}{rcl}f\left(8\right)& =& (8{)}^{1/3}\\ & =& 2\\ & & \end{array}$

Differentiate the function $x^{1/3}$to obtain the first derivative of the function.

$f^{\text{'}}\left(x\right)=\frac{1}{3}{x}^{-2/3}$

Substitute $8$for $x$in the above equation as

$\begin{array}{rcl}d{f}^{\text{'}}\left(8\right)& =& \frac{1}{3}(8{)}^{-2/3}\\ & =& \frac{1}{12}\\ & & \end{array}$

Differentiate the function$\frac{1}{3}{x}^{-2/3}$ to obtain the second derivative of the function.

$f^{\text{'}\text{'}}\left(x\right)=-\frac{2}{9}{x}^{-5/3}$

Substitute $8$for$x$in the above equation as

$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(8\right)& =& -\frac{2}{9}(8{)}^{-5/3}\\ & =& -\frac{1}{144}\\ & & \end{array}$

Differentiate the function $-\frac{2}{9}{x}^{-5/3}$ to obtain the third derivative of the function.

$f^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{10}{27}{x}^{8/3}$

Substitute $8$for$x$ in the above equation as

$\begin{array}{rcl}{f}^m\left(8\right)& =& \frac{10}{27}(8{)}^{8/3}\\ & =& \frac{5}{3456}\\ & & \end{array}$

Substitute all the values in the Taylor expression as follows:

$$\begin{gathered} f\left(x^{1/3}\right)=2+\frac{\frac{1}{12}}{1!}(x-8)+\frac{-\frac{1}{144}}{2!}(x-8{)}^2+\frac{\frac{5}{3456}}{3!}(x-8{)}^3+\dots \\ =2+\frac{1}{12}(x-8)-\frac{1}{288}(x-8{)}^2+\frac{5}{20736}(x-8{)}^3+\dots \end{gathered}$$

Thus, the first few terms of the Taylor series of $\sqrt[3]{x}$are:

$2+\frac{1}{12}(x-8)-\frac{1}{288}(x-8{)}^2+\frac{5}{20736}(x-8{)}^3+\dots$