Question

how do you find the continuity and discontinuity of a function?

Answer 1

an example would be good, but generally all non piecewise functions are continuous on their domain

Answer 2

but how do you find the points of discontinuity?

Answer 3

y = e^1/x

Answer 4

so for example answering "for what values of \(x\) is \(f(x)=\frac{x}{x-1}\) is identical to asking what the domain is

Answer 5

so it'd be x = 1?

Answer 6

well that wouldn't be the domain, it'd be ARN except for x=1

Answer 7

what is the domain of
\[f(x)=e^{\frac{1}{x}}\]?

Answer 8

I have no idea..

Answer 9

exponentials never made sense to me

Answer 10

yes, the answer to my question was all real numbers except 1
the answer to your question is identical to asking what the domain is,
what is the domain of \(\frac{1}{x}\)?

Answer 11

all real numbers except 0
how'd you get ARN except 1 for the e^1/x?

Answer 12

it is not a matter of exponentials in this case
the domain of \(e^x\) is all real numbers.
the problem is you still cannot divide by 0, so the domain of \(\frac{1}{x}\) is ARN except 0, and therefore so is the domain of \(e^{\frac{1}{x}}\)

Answer 13

i didn't, we must have crossed paths

Answer 14

i mean crossed posts. the answer to your question is all real numbers except 0, because exponentials or no, you still cannot divide by 0

Answer 15

simply put, writing
\[e^{\frac{1}{0}}\] makes no sense
so 0 is out of the domain, making the function necessarily discontinuous there

Answer 16

alright, so there's a discontinuity at 0,0?

Answer 17

at 0

Answer 18

then why does it ask for the points of discontinuity?

Answer 19

\(f(x)=e^{\frac{1}{x}}\) has no second coordinate at \(x=0\)
that is one reason why it is discontinuous
as x gets close to zero from the right (through positive numbers) \(\frac{1}{x}\) goes to positive infinity, which means \(e^{\frac{1}{x}}\) goes to positive infinity real real fast

Answer 20

"point" in this case is a synonym for real number (and \(x\) value). not a "point" in the plane

Answer 21

but isn't 0 the continuity? I don't understand the difference between the two then

Answer 22

intersting i had never thought of it that way before, but i can see the confusion

Answer 23

no zero is the point of DIScontinuity, it is continuous everywhere else

Answer 24

when will there be a continuity then?

Answer 25

lets look at a picture
http://www.wolframalpha.com/input/?i=e^%281%2Fx%29
you see that it is going to infinity (in the \(y\) direction of course, as \(x\) goes to zero.

Answer 26

it is continuous everywhere except at 0

Answer 27

as i said in the post above, almost all function that are written as equations, like
\[f(x)=\frac{x}{x-1}\] or
\[g(x)=x^2+x-1\] are continuous everywhere are their domains. means the graph has not breaks or holes
sine and cosine are continuous everywhere
all the rest of the trig functions are continuous on their domains
\(e^x\) is continuous everywhere
\(\log(x)\) is continuous on its domain \(x>0\)

Answer 28

*on there domains

Answer 29

then what isn't continuous? jump discontinuities?

Answer 30

if you see a piecewise function like
\[f(x) = \left\{\begin{array}{rcc}
1-x & \text{if} & x < -4 \\
x^2& \text{if} & x \geq-4
\end{array}
\right.
\]
then it will be continuous everywhere except possibly where the definition of the function changes, in this example at \(x=-4\)

Answer 31

your example above has an 'infinite" discontinuity, because at 0 is undefined, but going to infinity
the piecewise function i wrote above will have a "jump" discontinuity at \(x=-4\)

Answer 32

as \(x\) approaches \(-4\) from the right it is headed towards 5, but then at \(x=-4\) you get \((-4)^2=16\) and that is the jump, from 5 to 16

Answer 33

damn i meant from the left, not from the right, sorry

Answer 34

jump

Answer 35

that makes sense, so e^(1/x) is an infinite discontinuity?

Answer 36

infinite