Trying to understand some exposition on the 2nd derivative I struggle to understand non geometric (i.e. picture) based mathematical exposition, but I realize that I will need to understand it eventually. Part of my confusion lies in that I am not always sure what the symbols mean. I was trying to understand why the 2nd derivative test works, and after googling came across this website with the following statement:

Suppose that f does NOT have a local minimum at x = c. Then we cannot have the first derivative of f(x) negative on some interval (c - d,c) AND positive on some interval (c, c + d). Thus one of these two conditions fails.

My understanding of the symbols used are as follows:


*

*f is an arbitrary function

*x = c is referring to the y-axis being moved along the x-axis by some real number c

*d is also some arbitrary real number, i think


That being said, I do not understand the part of the statement which has negative on some interval (c - d, c), and being positive on some interval (c, c + d). What is (c - d, c) supposed to mean? Particularly, what is c - d in this context trying to tell me? Whats the best way to think about such statements?
 A: I take it that you are asking about the meaning of the interval $(c-d,c)$. Consider the point $c\in\mathbb{R}$. Let us draw the real line and select the point $c$. Put two sliders on this point, and move one by $d$ units to the left. The region between the two - not including $c$ and $c-d$ itself - forms the open interval $(c-d,c)$.
About the part where $f^{'}$ is negative on $(c-d,c)$ and $f^{}$ is positive on $(c,d+c)$. Drawing out the graph shows that in the interval $(c-d,c+d)$, the graph is locally a function of the form $f(x)=x^{2k}$. Also, you see that at the point $c$, the first derivative changes sign from $-$ to $+$. Using the graph shows that $f$ has a local maximum at $c$. This is the geometric interpretation of first derivatives, second derivatives, e.t.c.
A: Interval $(c-d, c)$ is the open interval $\{x \mid c-d < x < c \}$.  That is, all values of x between (but not including) the values $c-d$ and $c$.  Likewise $(c, c+d)$ is the open interval $\{x \mid c < x < c+d\}$.
If the curve is continuous and has no local minimum at $x=c$ then the gradient of the curve (the derivative) must have the same sign on either side near that point.  It cannot have a negative slope on one side and a positive slope on the other.
(Of course this only works for arbitrarily small $d$; sufficiently so that there are no minima in either intervals.)

It's more conventional to separate the bounds by double dots, like so $(a..b)$, to be clear that you're describing an interval rather than a coordinate pair (or a series).
For a closed interval, one uses 'square-braces' on the closed end. Such as $[a..b) \equiv \{x | a \leq x < b\}$.
