Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x^3-1)(x^2-x+1)\\
&=(x-1)(x^2+x+1)(x^2-x+1).
\end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$? 
 A: The derivative has a single real root, namely $1$. The other two factors never vanish on $\mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-\infty$ and $\infty$ are both $\infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}=\frac{30-20+15-12+10}{60}=\frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
A: egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $\ln(x+1)$:
$$\ln(x+1)=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-\ln(x+1)+O(x^7)+1.$$
As $x\to-1$ (here, the series expansion works), $\ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $x\to\infty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
