# Peicewise Functions Continuity

• Aug 26th 2009, 05:46 PM
The Power
Peicewise Functions Continuity
I am doing some review, starting back at the fundamentals and I do not know why but piecewise confuse the hell out of me at times. I have the following piecewise function and have to tell if it is continuous or not on the interval [-1,1]

$\displaystyle f(x)=\frac{x}{|x|} x not equal to 0$

$\displaystyle f(x)= x = 0 when x = 0$

Sorry to be so abrupt, I know how to tell continuity with normal functions such as if it a rational fraction and long as the value that deems the function undefined is not in that interval it is continuous? also if the sign changes from left to right in a interval?

Any tips to help me with continuity
• Aug 26th 2009, 07:15 PM
math84
Quote:

Originally Posted by The Power
Any tips to help me with continuity

Graph it out and make sure it doesn't have any gaps
• Aug 26th 2009, 07:51 PM
The Power
We are told to evaluate it algebraically since we are not allowed to use calculators on a test. Thanks for the help though, it know its not terrible hard, its just that piecewise functions to through me off at times.

Edit: I think I finally recall my information and remove this mental block. A piecewise function is a whole function just has different rules for set intervals. So in sense if the interval for
$\displaystyle f(x)=\frac{x}{|x|}$ when x is not equal to 0,

Sorry I forgot latex tags. So if the rule was a tad different and allowed 0, this would cause the function to be undefined causing a &quot;gap&quot; or hole in the graph thus rendering it no longer continuous, however since we have the rule when x equals 0 the output is 0 this allow the graph to continue on the interval [-1,1] Please someone correct me if I am wrong.
• Aug 27th 2009, 02:49 AM
Plato
$\displaystyle \frac{x}{{\left| x \right|}} = \left\{ {\begin{array}{rl} {1,} & {x > 0} \\ { - 1,} & {x < 0} \\ \end{array} } \right.$
• Aug 27th 2009, 03:16 AM
HallsofIvy
As Plato said, if x< 0, |x|= -x so $\displaystyle \frac{|x|}{x}= -1$ for x< 1. If x> 0, |x|= x so $\displaystyle \frac{|x|}{x}= 1$ for x> 1.

In order for the function f(x) to be continuous at x= a, three things must be true:
1) f(a) is defined
2) $\displaystyle \lim_{x\rightarrow a} f(x)$ exists
3) $\displaystyle \lim_{x\rightarrow a} f(x)= f(a)$
(Since $\displaystyle \lim_{x\rightarrow a} f(x)= f(a)$ pretty much implies the two sides exist, usuallly we just state (3).)

Now, if $\displaystyle \lim_{x\rightarrow a} f(x)$ exists then the two "onesided limits", $\displaystyle \lim_{x\rightarrow a^-} f(x)$ and $\displaystyle \lim_{x\rightarrow a^+} f(x)$ must exist and be equal. Think about what that means here.