Answer
False
Work Step by Step
By Definition 9.7.3, the $n$th Maclaurin polynomial for $f$ is given by
${p_n}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$
Thus, ${p_4}\left( x \right)$ for ${{\rm{e}}^x}$ is
${p_4}\left( x \right) = 1 + x + \dfrac{1}{{2!}}{x^2} + \dfrac{1}{{3!}}{x^3} + \dfrac{1}{{4!}}{x^4}$
${p_4}\left( 2 \right) = 1 + 2 + \dfrac{1}{{2!}}{2^2} + \dfrac{1}{{3!}}{2^3} + \dfrac{1}{{4!}}{2^4} = 7$
Therefore,
$\left| {{{\rm{e}}^2} - {p_4}\left( 2 \right)} \right| \approx 0.389056$
Since $\dfrac{9}{{5!}} = 0.075$, we have
$\left| {{{\rm{e}}^2} - {p_4}\left( 2 \right)} \right| \approx 0.389056 \gt \dfrac{9}{{5!}}$
Thus, the statement is false.
