[SPLIT] determining if discontinuity is removeable or not

crzymath

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Oct 27, 2008
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how do you know when a graph is continuos or if a discontinuity is non-removable or removable? this is my problem:

f(x)= { x[sup:2axelbfj]2[/sup:2axelbfj]+ 1, x>=3
{9, x <3

says to find all the x values where x is not continuous and classify it.

another problem i have is sketching a graph with given conditions. for example, i will be given:

f(-1) = 1, f(1) = -1, f(2) = -4, f(4) = 0, f(3) = 2;
f '(1) = 0, f '(-1) and f '(3) are undefined;
f '(x) > 0, for x < -1 and 1 < x < 3;
f '(x) < 0 for -1 < x < 1 and x > 3;
f ''(x) > 0 for x < -1 and x > -1;

what would i do with these statements?
 
Check this for a nice definition: http://www.mathwords.com/r/removable_discontinuity.htm

Look to see if yours has a discontinuity at x=3. It does not appear to be removable.

If we check the limits as x approaches 3 from the left and right, we get different values. Thus, the discontinuity is not removable. This appears to be what is called a 'jump discontinuity'. Notice how it jumps from 9 up to 10, at 3

\(\displaystyle \lim_{x\to 3^{-}}f(x)=9\)

\(\displaystyle \lim_{x\to 3^{+}}f(x)=10\)

A removable discontinuity would be something like this:

\(\displaystyle \frac{x^{2}-4}{x^{3}-8}\).

Let's factor and we get \(\displaystyle \frac{(x+2)(x-2)}{(x-2)(x^{2}+2x+4)}\)

See the x-2?. It cancels. That is the hole or removable discontinuty.
 
crzymath said:
... another problem i have is sketching a graph ...

... f(-1) = 1, f(1) = -1, f(2) = -4, f(4) = 0, f(3) = 2 ...

... what would i do with these statements?


Do you recognize these five points? You could start by graphing them.

 
ok that makes sense-graph those x,y points. but what about the rest of the conditions?? what good are they??
 
crzymath said:
... but what about the rest of the conditions? what good are they?


The rest of the conditions tell you enough about the behavior of function f to be able to sketch its rough graph.

 
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