Function

1. Yes, it is a discontinued function.
One distinct point at (0,0) and a continuous curve at [1,inf)2. No. y = ln x valid only for x belongs to (0,inf).
There is a lack of functioned values for x belonging to R\(0,inf) = (-inf,0]3. No.
(i) Lack of functioned values for R\[-1,1]
(ii) More than one* functioned values for each x within [-1,1]Remarks:
* This depends on the definition of f(x).What we have to bear in mind is that, a function must give one and only one functioned value within the range for every x which belongs to the domain. .

Supplementary Information:

..For each function, domain must be "filled up" (Each x has a y assigned)
..But the range does not needed to be "filled up".
..The set of functioned value is called "co-domain", which must be a subset of range.

..For Question 1, the function is defined explicitly and the domain is nothing but {0}U[1,inf) the range is R.

..For Question 2 and 3, it involves the concept of Inverse Function.
..For a function to have inverse, it must be bijective, which means
..(i) one-one (injective); and
..(ii) onto (surjective).
..exp(x) is bijective and sin(x) is not.

學問 ( 中學級 1 級 ) : 你好。
.1.Yes, it is a discontinued function.
One distinct point at (0,0) and a continuous curve at [1,inf)
Your answer is wrong.
The fuction f(x) =x^1/2 *cos (1/x) foe x >=1 and 0 for x=0 is a continuous function。
Please refer to the definition of continuous function in WIKI

2011-09-08 00:20:32 補充：
The function f is continuous if it is continuous at every point of its domain. If the point c in the domain of f is not a limit point of the domain, then this condition is vacuously true, since x cannot approach c through values not equal c. Thus, for example, every function whose domain is the set

2011-09-08 00:21:43 補充：
http://en.wikipedia.org/wiki/Continuous_function
**Definition in terms of limits
The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through domain of f exists and is equal to f(c).