Showing that $x=0$ is a minimum point by using Taylor's expansion 
Find all the triplets $(\alpha ,\beta, \gamma)\in\mathbb{R}^3$, such that the function $$f(x)=e^x +\alpha \arctan(x) + \beta \arctan^2(x) +\gamma \arctan^3(x)$$
  has a minimum in $x=0$.

My solution.
By taking the derivative I obtain $$f'(x)=e^x +\frac{\alpha}{1+x^2} +\frac{2 \beta \arctan(x)}{1+x^2} +\frac{3 \gamma \arctan2(x)}{1+x^2}$$
So the limit $\lim_{x\to 0} f'(x)=0$.
Expanding the terms I find that every term has a $x$ beyond, except $1+ \alpha$, so the triplets are infinite $(-1, \beta, \gamma)$. 
Is this possible?
 A: Consider the Taylor expansion of $f$ at $x=0$ up to "some" order and recall the higher-order derivative test.
Up to the first order,
$$f(x)=1+x+\alpha x+O(x^2)=1+(1+\alpha) x+O(x^2)$$
which implies that if there is a minimum at $0$ then $1+\alpha=0$ (otherwise $f$ is strictly monotonic in a neighbourhood of $0$), hence $\alpha=-1$. 
Now, with $\alpha=-1$, up to the second order, 
$$f(x)=1+x+\frac{x^2}{2}+(-1)(x)+\beta(x^2)+O(x^3)=1+\left(\frac{1}{2}+\beta\right) x^2+O(x^3)$$
and if $\frac{1}{2}+\beta> 0$ then $0$ is a minimum point. It is a maximum point when $\frac{1}{2}+\beta<0$.
To check the case when $\frac{1}{2}+\beta=0$, that is $\beta=-\frac{1}{2}$, you should consider the expansion up to the third order:
$$f(x)=1+\left(\frac{1}{2}+\gamma\right) x^3+O(x^4).$$
If there is a minimum at $0$ then $\frac{1}{2}+\gamma=0$ that is $\gamma=-\frac{1}{2}$. 
Finally, with $\alpha=-1$, $\beta=\gamma=-\frac{1}{2}$, in order to see if $0$ is a minimum point, check the expansion up to the fourth order.
Can you take it from here?
A: From $$f'(x)=e^x +\frac{\alpha}{1+x^2} +\frac{2 \beta \arctan(x)}{1+x^2} +\frac{3 \gamma \arctan2(x)}{1+x^2}$$
you find $f'(0)=1+\alpha $.
There is no  need for $ \lim_{x\to 0} f'(x)=0$
The other issue is that: how do you know that x=0 is a minimizer just by $f'(0)=0?$.
You need to verify that it is indeed a minimizer by second derivative test.
