Simple question right?
This website, along with the Calc book we’re teaching from, define it this way:
A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.
No debate about there being an inflection point at x=0 on this graph.
There’s no debate about functions like 
In fact, I think we’re all in agreement that:
- There has to be a change in concavity. That is, we require that for
we have
and for
we have
, or vice versa.*
- The original function
has to be continuous at
. That is,
does not have a point of inflection at
, we still don’t consider this a point of inflection because of the lack of continuity.
The point of inflection x=0 is at a location without a first derivative. A “tangent line” still exists, however.
But the part of the definition that requires ![f(x)=\sqrt[3]{x}](http://s0.wp.com/latex.php?latex=f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D&bg=ffffff&fg=333333&s=0 "f(x)=\sqrt[3]{x}")
Is x=0 a point of inflection? Some definitions say no, because no tangent line exists.
So It’s clear that this definition is built to include vertical tangents. It’s also obvious that the definition is built in such a way as to exclude cusps and corners. Why? What’s wrong with a cusp or corner being a point of inflection? I would claim that the piecewise-defined function 
I prefer the definition:
A point where the graph of a function is continuous and where the concavity changes is a point of inflection.
That is, I would only require the two conditions listed at the beginning of this post. What do you think?
Once you’re done thinking about that, consider this strange example that has no point of inflection even though there’s a concavity change. As my colleague Matt suggests, could we consider this a region of inflection? Now we’re just being silly, right?
A region/interval of inflection?
———————————————
Footnotes:
* When we say that a function is concave up or down on a certain interval, we actually mean
** This source, interestingly, seems to require differentiability at the point. I think most of us would agree this is too strong a requirement, right?
.
![\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_a^{1}\frac{1}{x}dx\right]](http://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7B-a%7D%5Cfrac%7B1%7D%7Bx%7Ddx%2B%5Cint_a%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0 "\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_a^{1}\frac{1}{x}dx\right]")
![= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}](http://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cln%28a%29-%5Cln%28a%29%5Cright%5D%3D%5Cboxed%7B0%7D&bg=ffffff&fg=333333&s=0 "= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}")
![\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_{2a}^{1}\frac{1}{x}dx\right]](http://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7B-a%7D%5Cfrac%7B1%7D%7Bx%7Ddx%2B%5Cint_%7B2a%7D%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0 "\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_{2a}^{1}\frac{1}{x}dx\right]")
![= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}](http://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cln%28a%29-%5Cln%282a%29%5Cright%5D%3D%5Cboxed%7B%5Cln%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D&bg=ffffff&fg=333333&s=0 "= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}")
![\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^-}\left[\int_{-1}^{a}\frac{1}{x}dx\right]+\lim_{b\to 0^+}\left[\int_b^{1}\frac{1}{x}dx\right]](http://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E-%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7Ba%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D%2B%5Clim_%7Bb%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_b%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0 "\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^-}\left[\int_{-1}^{a}\frac{1}{x}dx\right]+\lim_{b\to 0^+}\left[\int_b^{1}\frac{1}{x}dx\right]")
![= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]](http://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E-%7D%5Cleft%5B%5Cln%7Ca%7C%5Cright%5D%2B%5Clim_%7Bb%5Cto+0%5E%2B%7D%5Cleft%5B-%5Cln%7Cb%7C%5Cright%5D&bg=ffffff&fg=333333&s=0 "= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]")
and the second is 
and the overall expression has the indeterminate form 
. In our very first approach, we assumed the limit variables 
and 
were the same. In the second approach, we let 
. But one assumption isn’t necessarily better than another. So we claim the integral diverges.
. For goodness sake, it’s symmetric about the origin!
between -1 and 1, and then throw darts at the graph uniformly, wouldn’t you bet on there being an equal number of darts to the left and right of the y-axis?
and 
. Then 
and 
. Rewriting the integral and evaluating, we find
.
and compute all its antiderivatives. Multiply, then insert alternating signs and voila! In this case, we choose 
and 
. The result is shown below.
. Checking with 
.
