-A function is discontinuous at a if it is defined near a but not continuous at a. To prove that a function is discontinous, we must show which of the requirements of continuous functions that it fails to hold for.
-Removable discontinuity: It is called removable discontuniuity because the discontinuity can be removed by redefining the function so that it is continuous at a. Example: lim (x^2-10x+21)/(x^2+x-12) x>3 = (x-7)(x-3)/(x-3)(x+4) (x-3) cancels = (x-7)/(x+4) = (3-7)/(3+4) = -4/7 Can't plug 3 directly into equation becasue the denominator would equal 0, so instead you factor and towards the end you plug in 3. Which this function has a removable discontinuity at x=3
-Jump discontinuity: It is called jump discontinuity because the function jumps from the left-hand limit to the right-hand limit at each point. Example f(x) = -2x, x2 -2(2) = -4 2^2-4(2)+1 = -3 These x-values are not equal so therefore this function is a jump.
-Vertical Asymptotes: When finding this there are only 3 possible limits, which are infiniti, negative infiniti, or DNE. Example f(x) = -1/x-4 left: negative infiniti right: negative infiniti To find these, you draw a number line and then plug in numbers on each side to get your limits approaching from the left and right.
-Website used http://www.nipissingu.ca/calculus/tutorials/limits.html
Okay, you will normally find a jump in a piecewise. They are incredibly easy to find in my opinion, in fact, they are the easiest discontinuity to find out of the three. -To find a jump in a piecewise you simply plug in the x values, and if the equations do not equal the same thing, then there is a jump. Ex: ) x + 2 x > 1 > ) 2x x > 1 When you plug into the two equations, you get (1) + 2 = 3 for the first one and 2 (1) = 2 for the second one. The values are not the same, therefor there is a jump.
To find a removable you look at a fraction and factor it out. Then you cancel. If you set what you cancelled equal to zero and solve for x you will get your removable discontinuity. Ex: X + 2 / x^2 – 4 -factor the denominator using difference of two squares: X + 2 / (x + 2) (x – 2) -now the x + 2’s cancel, so set x - 2 equal to zero and solve…..that gives you x = 2 which is where the removable is…..wallah!!
Last but not least are the vertical asymptotes. These are very similar to finding removables….you factor, then cancel, then set whats left equal to zero. Ex: X – 4 / x^2 – 16 -factor X – 4 / (x + 4) (x – 4) X – 4’s cancel 1 / x + 4 -set x + 4 equal to zero and you get that there is a removable asymptote at x = -4
There is only one way to get a jump, which is in a piecewise. When you are given a piecewise, you take your x value and plug it into your function that they give you. If they come out to be equal then it is a jump. Ex: x+1 x < 0 =1 x^2 +1 x > 0 =1 When you plug in zero for both you get 1 so it is a jump.
There is two times when you can get a removable to be your answer which is in a piecewise and a fraction. When you’re given a fraction, you first need to try and factor the top and bottom, if you can cancel then you take what you canceled and set it equal to zero and solve for x. what x is equal to is a removable. The other time you can have a removable is in a piecewise, you take your x value and plug it into your function that they give you. If they come out not equal to each other and there is no less than or greater than equal to sign then it is a removable. Ex: x^2-1/x -1 factor: (x+1)(x-1)/(x -1. The x-1’s cancel so set equal to zero. x- 1= 0. x=1 is a removable.
For vertical asymptotes one way you can get one is with trig functions: Tan (x), cot (x), csc (x), and sec (x) When your trying to find the discontinuity and your given a fraction you have to try and factor the top and bottom and cancel if possible. Whatever is left after you cancel or if you can’t cancel you set that equal to zero and whatever x is equal to is a vertical asymptote. Ex: in the same problem as above you would take the x+1 and set it equal to zero and solve for x. because you have two parts to it. x+1=0. x=-1 is a vertical asymptote
-A Jump is when the left and right hand limits exist but do not agree. You will normally find a jump in a piecewise.
Ex. {-2x, ≤2 {x^2-4x+1, >2
plug in 2 to both and you get -4 and 1. Therefore, there is a jump.
-For a removable you factor the top and bottom first. If possible, cancel. And last, you set what you canceled equal to 0. What ever your answer is is the removable.
Ex. f(x)=(|x-8|)/(x-8)
(x-8) can cancel so you set it equal to 0 and you have a removable at 8.
-Finding vertical asymptotes are just like finding removables. You factor, cancel, then set whats LEFT in the DENOMINATOR equal to zero.
Ex. f(x)=(1)/(x^2+1)=(1)/(x+1)(x-1)
You cant cancel anything so you set the bottom equal to zero and you get plus and minus 1.
a. the one way that you can get a jump is in a piecewise. after you solve the piecewise if both sides that you solved for are not equal then you are dealing with a jump.
here is an example
3x+2 x>2 8 3x x<2 6
the answers are not equal. therefore you have a jump!
b. for a removable at times you can have a removable in a piecewise mostly it happens in a fraction
example:
x-2/x^2-4
you can factor the bottom of the fraction and if possible thats what you should do
x-2/x+2 x-2
x-2 cancel and then you are left with x+2 which you set to 0
this is how you get a removable you would have a removable at -2
c. you can get a vertical asymptote in a fraction just as you can get a removable. the difference is you start off with the same steps and if you are unable to factor and cancel you set the bottom equal to zero and solve,
example:
x+2/x-2
in this case there would be a vertical asymptote at 2
*plug 5 into both of them...for the top one, you would get -10, and for the bottom one, you would get -17. They are not the same, so you end up with a jump.
A removable is normally found in fractions:
x-1/x^2-1
^^Factor this problem: (x-1)/(x-1)(x+1)
*You factor it, then cancel what you can. Then you set what ever you cancelled equal to 0. Then that is your removable. So for this problem, your removable is x=1.
Vertical assymptotes can also be found in fractions:
x+2/x^2-4
^^factor this problem: (x+2)/(x+2)(x-2)
*You would factor the problem, and cancel what you can. After you factor, what is left you set equal to 0. And that is your vertical assymptote. So in this problem, x=2 would be your vertical assymptote.
-A function is discontinuous at a if it is defined near a but not continuous at a. To prove that a function is discontinous, we must show which of the requirements of continuous functions that it fails to hold for.
ReplyDelete-Removable discontinuity: It is called removable discontuniuity because the discontinuity can be removed by redefining the function so that it is continuous at a.
Example:
lim (x^2-10x+21)/(x^2+x-12)
x>3
= (x-7)(x-3)/(x-3)(x+4)
(x-3) cancels
= (x-7)/(x+4) = (3-7)/(3+4) = -4/7
Can't plug 3 directly into equation becasue the denominator would equal 0, so instead you factor and towards the end you plug in 3. Which this function has a removable discontinuity at x=3
-Jump discontinuity: It is called jump discontinuity because the function jumps from the left-hand limit to the right-hand limit at each point.
Example
f(x) = -2x, x2
-2(2) = -4
2^2-4(2)+1 = -3
These x-values are not equal so therefore this function is a jump.
-Vertical Asymptotes: When finding this there are only 3 possible limits, which are infiniti, negative infiniti, or DNE.
Example
f(x) = -1/x-4
left: negative infiniti
right: negative infiniti
To find these, you draw a number line and then plug in numbers on each side to get your limits approaching from the left and right.
-Website used
http://www.nipissingu.ca/calculus/tutorials/limits.html
Okay, you will normally find a jump in a piecewise. They are incredibly easy to find in my opinion, in fact, they are the easiest discontinuity to find out of the three.
ReplyDelete-To find a jump in a piecewise you simply plug in the x values, and if the equations do not equal the same thing, then there is a jump.
Ex:
) x + 2 x > 1
>
) 2x x > 1
When you plug into the two equations, you get (1) + 2 = 3 for the first one and 2 (1) = 2 for the second one. The values are not the same, therefor there is a jump.
To find a removable you look at a fraction and factor it out. Then you cancel. If you set what you cancelled equal to zero and solve for x you will get your removable discontinuity.
Ex:
X + 2 / x^2 – 4
-factor the denominator using difference of two squares:
X + 2 / (x + 2) (x – 2)
-now the x + 2’s cancel, so set x - 2 equal to zero and solve…..that gives you x = 2 which is where the removable is…..wallah!!
Last but not least are the vertical asymptotes. These are very similar to finding removables….you factor, then cancel, then set whats left equal to zero.
Ex:
X – 4 / x^2 – 16
-factor
X – 4 / (x + 4) (x – 4)
X – 4’s cancel
1 / x + 4
-set x + 4 equal to zero and you get that there is a removable asymptote at x = -4
There is only one way to get a jump, which is in a piecewise. When you are given a piecewise, you take your x value and plug it into your function that they give you. If they come out to be equal then it is a jump.
ReplyDeleteEx: x+1 x < 0 =1
x^2 +1 x > 0 =1
When you plug in zero for both you get 1 so it is a jump.
There is two times when you can get a removable to be your answer which is in a piecewise and a fraction. When you’re given a fraction, you first need to try and factor the top and bottom, if you can cancel then you take what you canceled and set it equal to zero and solve for x. what x is equal to is a removable. The other time you can have a removable is in a piecewise, you take your x value and plug it into your function that they give you. If they come out not equal to each other and there is no less than or greater than equal to sign then it is a removable.
Ex: x^2-1/x -1 factor: (x+1)(x-1)/(x -1. The x-1’s cancel so set equal to zero. x- 1= 0. x=1 is a removable.
For vertical asymptotes one way you can get one is with trig functions: Tan (x), cot (x), csc (x), and sec (x)
When your trying to find the discontinuity and your given a fraction you have to try and factor the top and bottom and cancel if possible. Whatever is left after you cancel or if you can’t cancel you set that equal to zero and whatever x is equal to is a vertical asymptote.
Ex: in the same problem as above you would take the x+1 and set it equal to zero and solve for x. because you have two parts to it. x+1=0. x=-1 is a vertical asymptote
-A Jump is when the left and right hand limits exist but do not agree. You will normally find a jump in a piecewise.
ReplyDeleteEx.
{-2x, ≤2
{x^2-4x+1, >2
plug in 2 to both and you get -4 and 1. Therefore, there is a jump.
-For a removable you factor the top and bottom first. If possible, cancel. And last, you set what you canceled equal to 0. What ever your answer is is the removable.
Ex.
f(x)=(|x-8|)/(x-8)
(x-8) can cancel so you set it equal to 0 and you have a removable at 8.
-Finding vertical asymptotes are just like finding removables. You factor, cancel, then set whats LEFT in the DENOMINATOR equal to zero.
Ex.
f(x)=(1)/(x^2+1)=(1)/(x+1)(x-1)
You cant cancel anything so you set the bottom equal to zero and you get plus and minus 1.
There are vertical asymptotes at ±1
a. the one way that you can get a jump is in a piecewise.
ReplyDeleteafter you solve the piecewise if both sides that you solved for are not equal then you are dealing with a jump.
here is an example
3x+2 x>2 8
3x x<2 6
the answers are not equal.
therefore you have a jump!
b. for a removable at times you can have a removable in a piecewise
mostly it happens in a fraction
example:
x-2/x^2-4
you can factor the bottom of the fraction and if possible thats what you should do
x-2/x+2 x-2
x-2 cancel
and then you are left with x+2
which you set to 0
this is how you get a removable
you would have a removable at -2
c. you can get a vertical asymptote in a fraction just as you can get a removable.
the difference is you start off with the same steps and if you are unable to factor and cancel
you set the bottom equal to zero and solve,
example:
x+2/x-2
in this case there would be a vertical asymptote at 2
A jump is normally found in a piecewise:
ReplyDelete{-2x, ≤5
{x^2-8x-2, >5
*plug 5 into both of them...for the top one, you would get -10, and for the bottom one, you would get -17. They are not the same, so you end up with a jump.
A removable is normally found in fractions:
x-1/x^2-1
^^Factor this problem: (x-1)/(x-1)(x+1)
*You factor it, then cancel what you can. Then you set what ever you cancelled equal to 0. Then that is your removable. So for this problem, your removable is x=1.
Vertical assymptotes can also be found in fractions:
x+2/x^2-4
^^factor this problem: (x+2)/(x+2)(x-2)
*You would factor the problem, and cancel what you can. After you factor, what is left you set equal to 0. And that is your vertical assymptote. So in this problem, x=2 would be your vertical assymptote.