Continuity or Discontinuity of a function?
Why is this the answeR?
because for a function to be continous at x three things must happen: 1) f(x) must be defined 2) the Limit of f(x) must exist 3) the Limit of f(x) must be equal the defined f(x) since the Limit as (x) approaches 1 from the left is different from the limit as (x) approaches 1 from the right, the Limit does not exist. Which means it must be discontinuous. Also the Limit as (x) approaches 1 from the left does not match the defined value of f(1), it must be discontinuous. all other points on the graph pass the three conditions, so it has to be discontinuous when x=1.
what are all the other points on the graph .. i dont understand? ... how do you know that it is discontinuous at x=1...why isnt it discontinuous at x = 2
alright lets test x=2 1) is f(2) defined? yes because f(2) = 10- (2) = 8 2) does the limit as x approaches 2 exist? yes because: coming from the left hand side\[\lim_{x \rightarrow 2} f(x) = 10 - (2) = 8\] coming from the right hand side\[\lim_{x \rightarrow 2} f(x) = 6(2) - (2)^2 = 12- 4 = 8\] so \[\lim_{x \rightarrow 2} f(x) = 8\] 3) does the limit match the defined value of f(2)? yes because they both are equal to 8 \[\lim_{x \rightarrow 2} f(x) = 8 \] \[f(2) = 8\] so the function has to be continuous at x=2 I hope that helped
Do the way the arrows point indicate which side it is coming from?
no the arrows always point left-to-right. Usually there will be a little minus sign (-) if its coming from the left, and a little plus sigh (+) if its coming from the write. I just had no idea how to put those in to the chat box haha.
oh i meant the greater than and less than signs.. sorry
oh well the greater and the less than signs are just telling you the interval to start using a new formula for f(x) so x < 1 is just telling you that all (x) left of 1, use f(x) = x^2 + 3 so if you needed to find the limit from the left hand side, then you use f(x) = x^2 + 3, if you needed to find the limit coming from the right hand side, then you use f(x) = 10 -x so yea you're right, in a way, they tell you when to use which formula for finding the limit.
okie dokie thats as best as i will get it .. im not a fan of continuous functions because i dont understand them very well
but thank you
Join our real-time social learning platform and learn together with your friends!