Answer
We prove that $f\left( x \right) = {{\rm{e}}^{ - x}}$ is represented by its Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}$ for all $x$.
Work Step by Step
From Table 2 we have
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
In Example 3 it is shown that ${{\rm{e}}^x}$ is represented by the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$.
Substituting $-x$ for $x$ in the series above gives
$f\left( x \right) = {{\rm{e}}^{ - x}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}$
Therefore, $f\left( x \right) = {{\rm{e}}^{ - x}}$ is represented by its Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}$ for all $x$.
