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To start at the start, my maths textbook says that:

  • A line, $f(x)$, is concave when $f''(x) ≤ 0$ (second derivative of $f(x)$ is smaller than or equal to zero, or the gradient of $f(x)$ is changing at a decreasing rate)
  • A line, $g(x)$, is convex when $g''(x) ≥ 0$ (second derivation of $g(x)$ is greater than or equal to zero, or the gradient of $g(x)$ is changing at an increasing rate)
  • A point of inflection is a point where the line changes from concave or convex or vice versa (the sign of the second derivative changes from positive to negative or vice versa)

Obviously this stirs up some confusion because on a line, $h(x)$, where $h''(z) = 0$, at the $x$-value of $z$, would be concave as well as convex (which to my understanding are mutually exclusive but apparently not, maybe). But then going through the some questions in the textbook, these definitions (of concave and convex) were proven to some extent (though some confusion still remains).

The first example we came across in class was the following:

$$f(x) = (x-5)^4$$

Find the coordinates of any point(s) of inflection.

So working it out would go as following: $$f'(x) = 4(x-5)^3\\\\ f''(x) = 12(x-5)^2\\\\ f''(x) = 0\\\\ 12(x-5)^2 = 0\\\\ (x-5)^2 = 0/12 = 0\\\\ x-5 = 0\\\\ x = 0+5 = 5\\\\ x = 5\\\\ y = f(5) = (5-5)^4 = 0^4 = 0 $$ Coordinates of point of inflection: $(0,0)$.

However, from knowing what a quartic line looks like, we known that there are no points of inflection. For example, the line $j(x) = x^4$ , we already known resembles a (positive) quadratic line (essentially) in that it is has no point(s) of inflection and is entirely convex for all values of x. Well, $$j(x-5) = (x-5)^4 = i(x),$$ from the question above, is nothing more than a translation of $j(x)$ to $5$ units to the right. Therefore, $i(x)$ would keep the characteristics of $j(x)$ such as having no point(s) of inflection and being entirely convex for all values of $x$.

Therefore, even though the second derivative of $i(x)$ equals zero it is not a point of inflection because the line does not change from convex to concave at this point. Instead, as with the rest of the line, it is just another point on the (entirely) convex line. Though this doesn't seem to be a very mathematical conclusion, at least not algebraic but rather a logical assumption.

This suggests that (at a particular value of $x$) the second derivative can equal zero and still be a point on a convex line (rather than being a point of inflection). I assume the same applies for a concave line, for example, $$k(x) = -(x-5)^4.$$

This supports the claim that a convex line can have a gradient of zero and still be concave or convex.

So the way that the book recommends to figure out whether a coordinate is a point of inflection or just another point on a concave or convex line (when the second derivative equals zero) is to measure the second derivative of an $x$-value just before and just after the coordinate you have just found to have a second derivative equal to zero. If one is positive and the other is negative (order is irrelevant) then it is a point of inflection.

I do not like this method because if there were another point of inflection (which you are unaware of due to lack of full working or some unknown, algebraic variable) that happens to be very close to the coordinate that you have just found to have a second derivative of zero (assuming coordinate you have just found is also a point of inflection) then the second derivatives that you measure just before and just after may be the same (if you have skipped over the second point of inflection) implying that it is not a point of inflection when in reality it is.

So this is where my remaining confusion lays:

1. Can a point on a line (where the$x$-value's second derivative equals zero) ever be concave AND convex (rather than just the concave or convex example presented above)?

2. Is there an algebraic way of analyzing the second derivative and deciding whether it is actually a point of inflection or not (rather than just looking at the original line which may no always be available)?

For the first question, my assumption is no, but I'm not sure so I'm asking.

For the second question, my assumption is that there some sort of relationship between the second derivative and whether or not that point is a point of inflection. This is because in the example above, $$i''(x) = 12(x-5)^2.$$ I have a feeling that the even power of the function of $x$ has something to do with reversing the change of second derivative's sign from positive to negative (or vice versa), as does the even power applied to a negative number reverse the negative number's sign, for example, $$(-4)^2 = 16.$$ Perhaps the fact that the value $5$ for $i''(x)$ would give zero twice over (again, due to the $x-5$ being the squared) has something to do with the fact that the point is in fact not a point of inflection.

I've also come across similar examples as $i(x)$ using the second derivative for trigonometric function. I don't believe any powers were involved in the function's second derivative (at least none that were even) which may debunk the first theory that an even power of the function of $x$ in a second derivative is what prevents the point (where the second derivative equals zero) from being an actual point of inflection. However, the second theory that a value of $x$ could be given an even number of times over to give prevent the point (where the second derivative equals zero) from being an actual point of inflection, may still intact since it is common in trigonometric functions for several x$x$-values to correspond to a single $y$-value or derivative value or second derivative value.

I would have thought that only the second derivation mattered in whether something is an actual point of inflection or not, therefore any second integration of, for example, $i''(x)$ would give a line where a point of inflection lies where $x = 5$ whether the second integration led to (the original) $i(x)$ or something else. I tried it once or twice with some random number, but I could neither prove that the point where $x = 5$ was or was not a point of inflection, this is because the only way I could think of (even) attempting to support my theory was using my graphical calculator to draw out the line and looking at the line where $x = 5$, which is highly unreliable since I'm only judging it by eye. I also didn't want to measure the second derivative of the $x$-values just before and after the point on the line where $x = 5$ because that is somewhat contradictory and a method that I'm trying to see if I can avoid.

Alternatively, my teacher thinks that the even power of the original line, for example, in $$i(x) = (x-5)^4$$ is what prevents the point on the line where $x = 5$ from being a point of inflection, though I think this is fairly vague. It would make more sense that the second derivative (rather than the original line) would have the direct impact on whether the point is and actual point of inflection or not (where the second derivative of it $x$-value equals zero).

So I guess another few question arises being:

1. Will $p''(x) = a(x+b)^{2n+1}$ and $p''(-b) = 0$ always give a point of inflection where $x = -b$, and will $q''(x) = c(x+d)^{2n}$ and $q''(-d) = 0$ never give a point of inflection where $x = -d$?

2. And can it be proven algebraically?

3. Is there a different, direct relationship between a line's second derivative and whether a point on that line (such that its $x$-value's second derivative equals zero) is a point of inflection or not?

Thanks for bearing with me if you've read this far. Let me know if there's anything that needs to be better explained. Overall, there are 5 questions, answers to any and all questions are appreciated.

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    $\begingroup$ Yes, concave and convex are not mutually exclusive $\endgroup$ – Tomasz Tarka Feb 9 '18 at 2:13
  • $\begingroup$ Please use MathJax to format!! $\endgroup$ – Saad Feb 9 '18 at 2:41
  • $\begingroup$ @TomaszTarka really? Wow, wouldn't have expected that. Could you give me an example of a point on a line (or an entire line) than is both concave and convex (since they're not mutually exclusive) please (to help me better comprehend it). $\endgroup$ – Alex P Feb 9 '18 at 11:18
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    $\begingroup$ Concavity and convexity is property of a function on an interval, not of point. Simplest example of function that is both concave and conves is $f(x)=x$, or even simpler, $f(x)=const$. And I think you may want to work with more general definition of convexity and concavity, which firstly doesn't require function to be twice differentiable, and gives you more insight to what concave functions really are $\endgroup$ – Tomasz Tarka Feb 9 '18 at 16:53

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