For which values of c does a function have 2,1 or 0 inflection points

Question

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.

Answer 1

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.

Answer 2

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 $x=b$.