# Inflection points and differentiation

I was looking at a homework for a calc class I have online, couldn't find it, and searched online. I found this answer to these questions:

• What is the minimum number of inflection points that must exist between (and not at ) two critical points of a non-constant differentiable polynomial?

A: The minimum number of inflection points is 0, because the critical points may not be of the form max-min.

• What is the maximum number of inflection points that can exist between two critical points of a differentiable function?

A: Only one.

Both of these answers were incorrect. The justification for this was:

• The correct answer is 1 because if you have two critical points that means there is either 2 maximums, 2 minimums or 1 maximum and 1 minimum. In any of these cases there has to be at least 1 inflection point.

and

• The correct answer is "There is no maximum" because there can be endless number of inflection points. The concavity can change a million times between two critical points.

In addition, the question:

• Where does the function $R(x) = {(x^2-9)(x^2-6x+5)\over (x-3)(x^2-2x-3)(x-1)(x^2+4)}$ have an asymptote?

A. I got it wrong when answering x=1,3 instead of their answer of x=-1,3.

If you look at the graph it shows x=-1,1,3

Definition of Critical Point: Critical points are the places on a graph where the derivative equals zero or is undefined. Interesting things happen at critical points.

The question I have is: is the answer on StackExchange wrong, is the answer on the homework site wrong and how?

By thinking about it and drawing graphs I'm pretty sure your answer sheet is correct about the properties of real, polynomial functions (your first two questions). I have not been able to prove it, yet, but the answers seem likely to me.

Your last question, however, I can answer. Vertical asymptotes occur at points $x_0 \in \Bbb R$ such that $\lim_{x\to x_0^\pm} f(x) = \pm \infty$.

In a rational function, the only values $x_0$ where there may be a vertical asymptote are values in which the denominator is equal to zero. At such values $x_0$ if the numerator is nonzero, then you are guaranteed to have a vertical asymptote. At values where the numerator is also zero, you'll have to evaluate the limit to determine whether the function diverges to $\pm \infty$ there or not.

The places where the denominator of $R(x)$ is zero are at $x_0 \in \{-1,1,3\}$.

Of those, the only $x_0$ where the numerator is not also zero is $x=-1$. So this is a vertical asymptote.

Then if you evaluate the limits at $x=1$ and $x=3$, you'll see that $|\lim_{x\to 1} R(x)|\lt \infty$ (I didn't bother to completely evaluate the limit because I was able to just cancel the $x-1$ out of both the numerator and denominator) and $\lim_{x\to 3} R(x)= \infty$. Thus $x=3$ is your only other vertical asymptote.

(I can't comment yet so my sincerest apologies for writing my comments in regards to $R(x)$ here).

When finding the vertical asymptote(s) for a rational function it is best to factorize the numerator and denominator and cancel out the common terms.

So, $R(x)$ simplifies to $\frac{(x−5)(x+3)}{(x−3)(x+1)(x^2+4)}$. From here it is clear that the vertical asymptotes are $x=−1$ and $x=3$.

Also, note that we say that $x=a$ is a vertical asymptote of $R(x)$ if $\lim_{x\to a^{−}}R(x)=\pm\infty$ or $\lim_{x\to a^{+}}R(x)=\pm\infty$.

If you look at $x=1$, then $\lim_{x\to 1}R(x)=\frac{4}{5}$, so $x=1$ is not a vertical asymptote.