It's the same. Continuity is a local concept: a function is continuous on some domain D if it is continuous at every point in D. Since your f is continuous at every point in [tex]\mathbb{R}-\{0\}[/tex], it is continuous on every subset [tex]D\subseteq\mathbb{R}-\{0\}[/tex], in particular on [tex]D= (-\infty,0) \cup (0,\infty)[/tex].ronaldor9 said:For example, if [tex]f(x)=\frac{1}{x} [/tex] then on the interval [tex] (-\infty,0) \cup (0,\infty), f(x)[/tex] is continous? Or is the function [tex]f(x)[/tex] continuous on [tex] (-\infty,0) [/tex] by itself and [tex] (0,\infty)[/tex] by itself?
Yes, a function can still be continuous on a composed interval even if it has a discontinuity at some point within the interval. This is because continuity is evaluated at each point individually, so as long as the function is continuous at all other points within the interval, it can still be considered continuous on the composed interval.
Pointwise continuity refers to a function being continuous at each individual point within the interval, whereas uniform continuity refers to a function being continuous on the entire interval as a whole. In other words, for uniform continuity, the function's rate of change must not vary significantly across the entire interval, while for pointwise continuity, the function's rate of change may vary at different points within the interval.
Continuity and differentiability are closely related concepts. A function must be continuous at a point in order to be differentiable at that point. This means that if a function is not continuous at a point within the interval, it cannot be differentiable at that point.
Yes, a function can still be considered continuous on a composed interval even if it has a removable discontinuity at one of the endpoints. This is because the value of the function at the endpoint can be adjusted to make the function continuous on the entire interval. However, if the discontinuity is non-removable, then the function cannot be considered continuous on the composed interval.
No, a function can be continuous on a composed interval even if it is not continuous on both intervals. As long as the function is continuous on the points where the two intervals overlap, it can be considered continuous on the composed interval. However, if the function is not continuous on any of the points where the intervals overlap, then it cannot be considered continuous on the composed interval.