Question

pls explain if possible: At what point(s) (if any) is the following function continuous? (pls see attached< I am stumped with this one)
f(x)={-x+1 x= an integer
{(x-1)^2 otherwise

Answer 1

didnt i tell u

Answer 2

yes you did Him, I still don;t understand thou> I think the term 'otherwise' is confusing me.

Answer 3

Assuming f(x) is a one-real-parameter real-valued function, then, it is defined as -x+1 when x is integer (0,1,2,3...), and (x-1)^2 everywhere else.

As
\[-x+1 \cap (x-1)^2 = \left\{ (x=1, y= 0 ) \right\}\]
The function does not contain any gap : it is continuous.Gor reference, see
http://www.wolframalpha.com/input/?i=plot+ {-x%2B1+%2C+%28x-1%29^2}

Answer 4

a) Continuous at x = 0, -1 and all non-integral values
b) Continuous at x = 0, 3 and all non-integral values
c) Continuous at x = 0 and 3
d) Continuous at x = 0, 1 and all non-integral values
e) Continuous at x = 1 and all non-integral values

Answer 5

that makes sense.

Answer 6

so even at the point x=0 they will be continuous?

Answer 7

or only at the point x=1?

Answer 8

f(x) is continuous when:
- x= 1
- x isn't an integer

In other word : f(x) ins't continuous whenever x is an integer (-1,0,1,2,3), execpt when x=1