Answer
(a) ${p_3}\left( x \right) = 1 + 2x - {x^2} + {x^3}$
(b) ${p_3}\left( x \right) = 1 + 2\left( {x - 1} \right) - {\left( {x - 1} \right)^2} + {\left( {x - 1} \right)^3}$
Work Step by Step
(a) We take the derivatives of $f\left( x \right) = 1 + 2x - {x^2} + {x^3}$:
$f'\left( x \right) = 2 - 2x + 3{x^2}$, ${\ \ \ }$ $f'\left( 0 \right) = 2$
$f{\rm{''}}\left( x \right) = - 2 + 6x$, ${\ \ \ }$ $f{\rm{''}}\left( 0 \right) = - 2$
$f{\rm{'''}}\left( x \right) = 6$, ${\ \ \ }$ $f{\rm{'''}}\left( 0 \right) = 6$
By Definition 9.7.2, the third Maclaurin polynomial for $f$ is
${p_3}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{x^3}$
${p_3}\left( x \right) = 1 + 2x - {x^2} + {x^3}$
Thus, ${p_3}\left( x \right) = f\left( x \right)$.
(b) We take the derivatives of $f\left( x \right) = 1 + 2\left( {x - 1} \right) - {\left( {x - 1} \right)^2} + {\left( {x - 1} \right)^3}$:
$f'\left( x \right) = 2 - 2\left( {x - 1} \right) + 3{\left( {x - 1} \right)^2}$, ${\ \ \ }$ $f'\left( 1 \right) = 2$
$f{\rm{''}}\left( x \right) = - 2 + 6\left( {x - 1} \right)$, ${\ \ \ }$ $f{\rm{''}}\left( 1 \right) = - 2$
$f{\rm{'''}}\left( x \right) = 6$, ${\ \ \ }$ $f{\rm{'''}}\left( 1 \right) = 6$
By Definition 9.7.3, the third Taylor polynomial about $x=1$ for $f$ is
${p_3}\left( x \right) = f\left( 1 \right) + f'\left( 1 \right)\left( {x - 1} \right) + \dfrac{{f{\rm{''}}\left( 1 \right)}}{{2!}}{\left( {x - 1} \right)^2} + \dfrac{{f{\rm{'''}}\left( 1 \right)}}{{3!}}{\left( {x - 1} \right)^3}$
${p_3}\left( x \right) = 1 + 2\left( {x - 1} \right) - {\left( {x - 1} \right)^2} + {\left( {x - 1} \right)^3}$
Thus, ${p_3}\left( x \right) = f\left( x \right)$.
