Inflexion Point

Question

Find the number of inflexion points on $$f(x) = \frac{x^2-5x+4}{x^2+5x+4}$$

I graphed the function and got this:

As you can see, there is a clear maximum and a clear minimum point on the graph, but there doesn't appear to be a third inflection point. I know in this case that the inflection point, although technically is inclusive of the maximum and minimum points, isn't the maximum or minimum point as these maximum and minimum points were mentioned separately in other subparts of the question. The answer given is 1.

I decided to use a different approach - I found the second derivative of the function to be:

$$f''(x) = -\frac{20(x^3-12x-20)}{(x^2+5x+4)^3}$$

Equating this to zero gives $x^3-12x-20 = 0.$ I then plotted this on the same coordinate axis:

This indicates a clear inflection point at roughly x = 4. I believe this to be the inflection point the answer is referring to. By definition, from Wikipedia, an inflection point is a point on a curve at which the curve changes from being concave (concave downward) to convex (concave upward), or vice versa.This is obviously not the case looking at the graph:

Why wasn't I able to identify this inflection point just by looking at the graph? This point does not seem to fit the given definition of an inflection point - what am I missing?

Answer 1

You may not have seen it because of the scale you have chosen on your graph. Note that to the right of the minimum, the graph is increasing but tending towards the horizontal asymptote, so there must be an inflection to the right of the minimum point.

Answer 2

Are you seeing the inflexion point now!? :)

For a better quality of the image click here or on the image itself.

Analytically, if you solve for $f^{''}(x)=0$ then you will see that this equation has only one real root namely $x=2\sqrt[3]{2}+\sqrt[3]4$.

Answer 3

Two ways to detect the inflection point, not mentioned yet:

(1) Blow up: Instead of plotting $f(x)$, plot $20\cdot f(x)$. You'll see immediately that there's an inflection point.

(2) Look closer: From your graph it is for sure that $f'$ is increasing up to approximately up to $x=4$. Now have a close look: the average slope between $4$ and $5$ is about $0.3$, between $6$ and $7$ about $0.2$, hence $f'$ is decreasing here.