Answer
${x^4} = 16 + 32\left( {x - 2} \right) + 24{\left( {x - 2} \right)^2} + 8{\left( {x - 2} \right)^3} + {\left( {x - 2} \right)^4}$
Work Step by Step
We compute the derivatives of $f\left( x \right) = {x^4}$ and list them in the table below:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}&{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}}\\
0&{{x^4}}&{{x^4}}&{16}\\
1&{4{x^3}}&{4{x^3}}&{32}\\
2&{12{x^2}}&{6{x^2}}&{24}\\
3&{24x}&{4x}&8\\
4&{24}&1&1\\
5&0&0&0
\end{array}$
By Theorem 1 of Section 11.7, the Taylor series centered at $c=2$ is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}{\left( {x - 2} \right)^n}$
Using the table above, we get
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}{\left( {x - 2} \right)^n} = 16 + 32\left( {x - 2} \right) + 24{\left( {x - 2} \right)^2} + 8{\left( {x - 2} \right)^3} + {\left( {x - 2} \right)^4}$
Thus, the Taylor series of $f\left( x \right) = {x^4}$ centered at $c=2$ is
${x^4} = 16 + 32\left( {x - 2} \right) + 24{\left( {x - 2} \right)^2} + 8{\left( {x - 2} \right)^3} + {\left( {x - 2} \right)^4}$
