1
$\begingroup$

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:

enter image description here

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:

enter image description here

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:

enter image description here

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?

$\endgroup$
  • $\begingroup$ The first derivative is positive. If there was not an infection point, the second derivative would be positive, too. Can a function, whose first and second derivatives are positive, converge to $1$ for $x\to\infty$? $\endgroup$ – Michael Hoppe Jul 26 '16 at 13:07
  • $\begingroup$ I can see a change in concavity. Zoom down and you'll see it clearly. Anyway, if $p(x)$ is the cubic polynomial in the numerator of $f''(x)$, it's easy to check $p(4)=-4$, $p(5)=45$n hence $f''(x)$ changes sign between $4$ and $5$. $\endgroup$ – Bernard Jul 26 '16 at 13:24
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The graph continues to increase as it approaches the horizontal asymptote... $\endgroup$ – StopReadingThisUsername Jul 26 '16 at 13:05
  • $\begingroup$ Yes, but the gradient tends to zero as x gets large $\endgroup$ – David Quinn Jul 26 '16 at 13:07
  • $\begingroup$ The gradient increases to a value at the inflection point from being equal to 0 at the minimum, and after the inflection point, begins decreasing as the graph approaches the horizontal asymptote. Is that right? $\endgroup$ – StopReadingThisUsername Jul 26 '16 at 13:12
  • $\begingroup$ Yes that's right. The inflection point is a local maximum in the value of the gradient. $\endgroup$ – David Quinn Jul 26 '16 at 13:24
2
$\begingroup$

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 $x=2\sqrt[3]{2}+\sqrt[3]4$.

enter image description here

$\endgroup$
  • $\begingroup$ this is not a proof you will need the third derivative of $f(x)$ $\endgroup$ – Dr. Sonnhard Graubner Jul 26 '16 at 13:24
  • $\begingroup$ @Dr.SonnhardGraubner: Yes but the OP knows that the second derivative has a root around $x=4$. His problem is just with visualizing the inflexion point! :) $\endgroup$ – H. R. Jul 26 '16 at 13:26
  • $\begingroup$ @Dr.SonnhardGraubner Not at all, Descartes' rule shows that the numerator of $f''$ has exactly one positive root, counting multiplicities as well. So there must be a change of sign. $\endgroup$ – Michael Hoppe Jul 26 '16 at 18:19
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.