Inflection point of a function For what values of $a$ does the function $f(x)=e^x+ax^3$ have an inflection point? 
It is an old question in my mind and wanted to bring it out.
 A: $$
f(x) = e^x + ax^3.
$$
$$
f'(x) = e^x + 3ax^2.
$$
$$
f''(x) = e^x + 6ax.
$$
Suppose first that $a>0$.
Since $f''(x)\to\infty$ as $x\to\infty$ and $f''(x)\to-\infty$ as $x\to-\infty$, and $f''$ is everywhere continuous, there must be a point where $f''$ changes signs.  And since $f'''(x)= e^x+6a$ is everywhere positive, $f''$ must be a strictly increasing function and hence can have only one point where it is equal to $0$, and there it must change signs.
Therefore there is only one solution to $e^x+6ax=0$, and that is where the inflection point is.  I don't think there's a closed-form solution, but you could apply Newton's method to solve this equation.
If $a=0$, then $f''(x)= e^x$, and this is always positive, so there are no inflection points.
Now suppose $a<0$.  Since $e^x\to0$ as $x\to-\infty$ and $6ax\to\infty$ as $x\to-\infty$, we have $f''(x)\to\infty$ as $x\to-\infty$.  But $f''(x)\to \infty$ as $x\to\infty$ as well.  Therefore the behavior of $f''(x)$ as $x\to\pm\infty$ does not necessitate any changes in sign of $f''(x)$.  If $a$ is a long way below $0$, you get a steep line plunging downward faster than $e^x$ grows when $x$ is near $0$, and $f''$ will become negative at some point where $x>0$, and then will become positive again to the right of that point.  You'll have two inflection points.  I think Newton's method will find them.  But if $a$ is negative and closer to $0$, the line going below $0$ will not move fast enough to outweigh the $e^x$ that is growing, and the graph of $f''$ will never go below the $x$-axis.  How do we know which values of $a$ are far enough below $0$ that there will be two inflection points?  So far all I can suggest is numerical experiments.
A: $$f''x=e^x+6ax$$
now,
$f''x=0$ when $e^x=-6ax$   ------[eqn1]
fram the graphs of $y=e^x$ and $y=-6ax$ it can be said that a solution to this will surely exist if $a>0$ ,
but when $a<0$ we things get complicated,
if a is too small the line $y=-6ax$ may not meet $e^x$ at all,
if a is too large, the line will meet the curve twice,
so we need to find the borderline case, when the line just touched the curve, i.e it is  tangent to $e^x$
Let the line touch the curve at some $x=x_1$
slope of the line $y=-6ax$ is $-6a$
slope of $y=e^x$ at for some $x=x_1$  is $e^{x_1}$
so we have $-6a=e^{x_1}$
also since the line and the curve are touching we have,
$-6ax_1=e^{x_1}$
from this and previous results we have,
$-6ax=-6a$
this gives the solution either $a=0$ or $x=1$
putting x=1 in [eqn1] we get,
$e=-6a$ or $a =\frac{-e}{6}$
hence the complete and final solution is , 
$$a \in [0,\infty)\cup(-\infty ,\frac{-e}{6}] $$
A: Actually there is a very nice closed-form solution to this problem.
As shown $f^{\prime\prime}(x)=e^x+6ax$. And inflection points are then solution to equation $$e^x=-6ax.$$
This have an analytical solution using so-called Lambert W function, i.e., the inflection points are: 
$$x^\star = -W\left(\frac{1}{6a}\right)$$
From conditions on existence of Lambert function we know that if $\frac{1}{6a}\geq -\frac{1}{e}$, then $x\in\mathbb{R}$, otherwise $x\in\mathbb{C}$.
This leads to the same conditions on $a$ as Tomarinator mentioned, i.e., if $a\in(-\infty,-e/6]\cup(0,\infty)$, then $x\in\mathbb{R}$. For other $a$ (except $0$), $x\in\mathbb{C}$, where in both cases, $x^\star=-W\left(\frac{1}{6a}\right)$.
