Question: The function $f(x)=x^3$ has the derivative $3x^2$ which is equal to $0$ at $x=0$ but the function $f(x)=x^3$ has neither maxima nor minima at that point. How is this possible?

Answer: 
Integrate the power series:

\begin{align}
\int \log x\, dx &= \int \sum\_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^i}{i} \\
&= \sum\_{i=1}^{\infty} (-1)^{i+1} \int \frac{(x-1)^i}{i} \\
&= \sum\_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^{i+1}}{i(i+1)} \\
&= (x-1)\sum\_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^i}{i(i+1)} \\
&= (x-1)\sum\_{i=1}^{\infty} (-1)^{i+1} \Big(\frac{(x-1)^i}{i} - \frac{(x-1)^i}{i+1} \Big)\\
&= (x-1) \log x - \sum\_{i=2}^{\infty} (-1)^{i} \frac{(x-1)^{i}}{i} \\
&= (x-1)\log x + (\log x - x+1) \\
&= x \log x-x + C.
\end{align}
Just calculate $$\left(x\ln{x}-x\right)'.$$