How can I know whether the point is a maximum or minimum without much calculation? 
Find the maximum and minimum of this function and state whether they
  are local or global:
$$f: \mathbb{R} \ni x \mapsto \frac{x}{x^{2}+x+1} \in \mathbb{R}$$

\begin{align*} 
f'(x)&= \frac{-x^{2}+x}{\left(x^{2}+x+1\right)^{2}}\\
f'(x)&=0 \iff \frac{-x^{2}+x}{\left(x^{2}+x+1\right)^{2}}=0 \iff -x^2+x=0 \iff x(1-x)=0, \end{align*}
which gives $x_{1}=0, x_{2}=1$. Here comes the disturbing part, we need to know if these are maximum or minimum and for this we usually used the second derivative. But this would be soo exhausting, I don't even  want think of doing it. There must be an easier way and I remember someone here has even recommended me using monotony somehow.
But how can we do this here? Please do tell me, at home I got enough time to use second derivative but surely not in the exam : /
 A: Assume your $f'(x)=0$ at $x=x_0$. Then there are three cases:


*

*if $f'(x)<0$ around $x=x_0^-$ and $f'(x)>0$ around $x=x_0^+$ then it is a minimum.

*if $f'(x)>0$ around $x=x_0^-$ and $f'(x)<0$ around $x=x_0^+$ then it is a maximum.

*if $f'(x)$ does not change sign around $x=x_0^-$ and $x=x_0^+$ then it is a turning point (provided that the second derivative exists).


In your case:


*

*the second one is true for $x_0=1$. Hence, it is a maximum.

*the first case is true for $x_0=-1$. Hence, it is a minimum.


You can see it in the following figure as well

A: Your expression for the derivative is wrong, but I'll let you sort that out. What is important is that:


*

*$f(0)=0$;

*there are just two values of $x$ for which $f'(x)=0$, and one of them is positive, the other negative;

*$f(x)$ tends to zero as $x$ tends to $\pm\infty$;

*$f(x)$ has the same sign as $x$.


These facts are enough to prove the nature of the stationary points. You should draw a graph if it helps. (You certainly don't have to calculate the second derivative.)
A: Another way that sometimes works is to compare the values of the function at these critical points. Often you can do this without the need to evaluate the function (ok, here the evaluation is immediate) $$f(x_1)=\frac{0}{0^2+0+1}=0$$ and $$f(x_2)=\frac{1}{1^2-1+1}=1$$ So, certainly $f(x_1)>f(x_2)$ and hence in $x_1$ you have a $\max$ and in $x_2$ a $\min$ (be careful for saddle points). Now, check what happens at $\pm \infty$ (take the limit) to conclude whether the $\min$ in $x_2$ and the $\max$ in $x_1$ are local or global. Spoiler:

 they are global.

