How do I show that the function
fn(x) = 1 + x + (x^2)/2! + ... + (x^n)/n!
has no real roots when n is an even number,
and has one real root when n is an odd number?
Suppose that $\displaystyle f_n(x)$ has the property that for any even $\displaystyle n$ we have no real roots. We can easy show that $\displaystyle f_n(x)$ has property that is has exactly one real root. Note that $\displaystyle f'_n(x) = f_{n-1}(x)$ where $\displaystyle n-1$ is even, and so the derivative is non-vanishing for $\displaystyle f_n(x)$, thus it is an increasing odd polynomial, thus, it has exactly one root.
There is an elementary way to prove that $\displaystyle f_n(x) > 0$ for $\displaystyle n$ even, which will complete what you want to show. Using this we can prove this. Let $\displaystyle F(x) = \frac{x^n}{n!}$ then $\displaystyle F(x)\geq 0$ if $\displaystyle n$ is even. This mean $\displaystyle f_n(x) = F(x)+F'(x)+...+F^{(n)}(x) \geq 0$.
I will try to come up with an approximating solution to this problem also, since I like putting bounds on functions.
Here is an approximation-type proof. First if $\displaystyle x>0$ then certainly $\displaystyle f_n(x) > 0$ and if $\displaystyle x=0$ then $\displaystyle f_n(x)\not = 0$. Thus, if is safe to assume that $\displaystyle x<0$. Let $\displaystyle f(x) = e^x$ by Taylor's theorem it means for any $\displaystyle x<0$ we have $\displaystyle e^x - f_n(x) = \frac{e^y}{(n+1)!} x^{n+1}$ where $\displaystyle x<y<0$ because $\displaystyle f_n(x)$ is a Taylor $\displaystyle n$ degree polynomial. Thus, $\displaystyle f_n(x) = e^x - \frac{e^y}{(n+1)!}x^{n+1}$. But $\displaystyle e^y>0 \mbox{ and }x^{n+1} < 0$ since $\displaystyle n$ is even and $\displaystyle x<0$. Thus, $\displaystyle e^x - \frac{e^y}{(n+1)!}x^{n+1} > e^x > 0$. This means, $\displaystyle f_n(x) > 0$ for all $\displaystyle x\in \mathbb{R}$ and $\displaystyle n$ even.
Are you familar with Taylor's theorem because that is all you need to know.