Inflection points and differentiation

Question

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?

Answer 1

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.

Answer 2

(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.