Considering the curve:
$$
f(x) = x²
$$
Does the curve $f$ have a maximum in the open interval $-1 < x < 1$? A minimum?
I had a hard time interpreting this question. My first thought was:
Yes, it has a maximum. The maximum value in the open interval $-1 < x < 1$ is
$$
lim_{x -> 1} x² = lim_{x -> -1} x² = 1
$$
The minimum value is naturally
$$
f(0) = 0² = 0
$$
However, I started wondering if thinking this way about the problem is wrong. Would it be more accurate to consider the problem as the whole curve $f$ having a maximum and a minimum in $-1 < x < 1$? Thus, arguing that it does NOT have a maximum in the interval (or any maximums at all, for that matter) but a minimum in $x = 0$, seeing as
$$
f'(x) = 2x, 2x = 0 \rightarrow x = 0
$$
Which way would be the most natural way to interpret the question?
Best Answer
If the interval contained $-1$ and $1$, which it does not, then the curve would have what we call local maximums at the endpoints of the closed interval, when considering the curve restricted to the interval. But the interval is open, so there are no endpoints to consider.
So there is no real need to pin down the intent of the question. There are no local or global maximums, but one local and global minimum, at the point $(0, 0)$ as you note.