For which values of c does a function have 2,1 or 0 inflection points 
Find for which values of $c$ that $f(x)=x^4+cx^3+\frac{x^2}{24}$ has:

(a) Two inflection points
(b) One inflection point
(c) Zero inflection points



I only know that $$f''(x)12x^2+6cx+\frac{1}{12}$$
But that's all I know how to do, also the determinant of that is $9c^2-1$ but I do not know how to proceed.
 A: The function $f''(x)$ is a quadratic polynomial whose limit at $\infty$ is $\infty$. It is either 


*

*positive everywhere,

*positive everywhere except for at a single point $a$, or

*positive on intervals of the form $(-\infty,a)$ and $(b,\infty)$ with $a < b$ and negative on the interval $(a,b)$.


The discriminant (which you computed incorrectly) can be used to distinguish between these three cases.
In the first case $f$ is always concave up - no inflection points.
In the second case $f$ is concave up on $(-\infty,a)$ and $(a,\infty)$ separately so again it has no inflection points.
In the third case $f$ has an inflection point at $x=a$ and $x=b$.
A: For two inflection points, 
the discriminant$\triangle>0$
Since the discriminant is $9c^2-1>0$
$$c^2>\dfrac19$$$$c<-\frac13\mbox{ or }c>\frac13$$
For one inflection point, 
Discriminant $9c^2-1=0$
$$c=\pm\dfrac13$$
But, notice that at $c=\pm\dfrac13$, we get the same concavity on left side and right side.
So there is no $c$ for which $f(x)$ gives one inflection point.
