Question

limit question

Answer 1

a). If it is continuous on more than one interval, use U for union. You may click on the graph to make it larger.

Answer 2

what is the question?

Answer 3

x
Use interval notation to indicate where f(x) is continuous. If it is continuous on more than one interval, use U for union. You may click on the graph to make it larger.

Answer 4

@hartnn

Answer 5

at which point you need to find the limit ??

Answer 6

I am not sure, the question above is what it asked me to do

Answer 7

:/

Answer 8

its limit question

Answer 9

okk...at which all points , do you see that the graph breaks ?

Answer 10

yes removal? discounituniy? it breaks everywhere. it confuses me!

Answer 11

ok, first easy things, its not defined for x<-2 and x>4
right ?

Answer 12

mm yes

Answer 13

then from -2 to -1 ,i don't see any break or discontinuity.
the point of interest is x=-1 , where its value given = 4
could you notice that ?

Answer 14

um 4.. is it x=4 or y=4?

Answer 15

value =4 means 'y' value =4
at x= -1

Answer 16

ok , I see!

Answer 17

now see the value of function JUST BEFORE x=-1 and JUST after x=-1, are they approximately same ???

Answer 18

or in precise terms, are they both very near to y value of -1 ??

Answer 19

After -1 is near??

Answer 20

whats the value of function JUST BEFORE x=-1 ??

Answer 21

y=-1?

Answer 22

yes,
whats the value of function JUST AFTER x=-1

Answer 23

mmm -2 twice??

Answer 24

are you seeing the left part for AFTER ?

Answer 25

yes, I am. is it wrong? :/

Answer 26

oh zero?

Answer 27

yes, but the point here is that on both sides of x=-1, the function value is approximately same and is -1

Answer 28

hence the function is continuous at x=-1

Answer 29

what about x=1 ?

Answer 30

I see.Theres three point
1.y=1
2.y=-1
3.y=3

Answer 31

i see
1.y=0
2.y=-1
3.y=3
but the question to be asked is, whether the value is same on left and right side , both ? approximately.

Answer 32

for left, -1
for right 3
for both 1

is this right?? :/

Answer 33

for left, -1
for right 3
means its different for both sides :P
so, the function is not continuous at x=1 !
got this ?
what about x=0 ?

Answer 34

6?? for both!

Answer 35

actually the graph goes to infinity from both sides, but good observation :)
since its same value on both sides, its continuous at x=0

so, finally, your total graph id only discontinuous at x=1 only.
got this ?

Answer 36

Okay! so the answer should be (-1,1)?

Answer 37

do you know how to write interval notation ?

Answer 38

mmm not 100 %, do i have to use U?

Answer 39

yes..

Answer 40

ok.. now im confused again

Answer 41

see, its not defined for x< -2 right ? so,
we start with
\([-2,\)
and we will go till the function is continuous, which is till x=1
so, we put \([-2,1]\)
got this ?

Answer 42

Yes im writing note please hold!

Answer 43

ok, whats next step?

Answer 44

[-2,1]U[1,4?

Answer 45

oh,thats correct :)

Answer 46

[-2,1]U[1,4]

Answer 47

I entered it but it tells me wrong!

Answer 48

:/

Answer 49

....
try [-2,1)U(1,4]

Answer 50

it tells me wrong again

Answer 51

:O did we miss any point ?

Answer 52

I dont know, mmmmmmmmm..

Answer 53

ok, let me try again.
[-2,-1]U[-1,0)U(0,1]U[1,4]

Answer 54

hmm...that can also be written as
[-2,0)U(0,4]

Answer 55

if one of these two doesn't come out to be correct...then all i can say is i tried my best to help you.

Answer 56

No it didnt work but thanks for helping tho. i understand limit much better

Answer 57

:( sorry couldn't help.

Answer 58

no, thanks for your time

Answer 59

@tkhunny

Answer 60

Too much to read. What problem are we working on?

Answer 61

its top of the page, the graph and find interal for limit

Answer 62

Are you joking? This is all one problem?!

Answer 63

yes..

Answer 64

Okay, so write it. What's stopping you?

Using the horizontal axis as 'x', the ONLY difficulty I see is at x = 0. It's just not clear on the graph at x = 0. You're going to have to decide about the endpoints. If they are included or not, can they be continuous? x = -2 and x = 4

Answer 65

not continuous?

Answer 66

since it stopped the line

Answer 67

I don't like your question mark. Is it or isn't it?

Answer 68

its not!

Answer 69

There can be continuity at any endpoint actually IN the Domain. If you end with an open circle (the end point is not actually IN the Domain), then we do NOT have continuity.

Since x = -2 is in the Domain and the right-side limit appears to exist, we get to include x = -2 in our list of continuity.

Since x = -2 is in the Domain and the right-side limit appears to exist, we get to include x = -2 in our list of continuity.

Since x = 4 is in the Domain and the left-side limit appears to exist, we get to include x = 4 in our list of continuity.

Everything else is an interior point and requires that both left and right limits exist at x = c AND that both limits are EQUAL and x = c AND that both limits EQUAL f(c).

That very last part discards x = 1 and x = -1 and maybe x = 0, but I already said I couldn't tell what the graph was saying at x = 0.

Answer 70

wow, thanks for the explanation.
so the answer should be (2,4)

Answer 71

no -2..

Answer 72

How about \([-2,-1)\cup (-1,0)\cup (0,1)\cup (1,4]\)

You need all the pieces.

Why did you throw out x = -2? What part of the one-sided definition is violated?

Answer 73

Thank you, why zero is included?

Answer 74

I'm guessing at x = 0. It APPEARS that around x = 0 the graph exhibits asymptotic behavior and that the function has been redefined as f(0) = 6. This is not continuous.

Answer 75

oh isee! thank you.
can i ask one more question?

Answer 76

Find c such that the function
f(x)=x^2-6 \[f(x)=x^2-6 x lec \]

Answer 77

f(x) x^2-6 x<=c
8x-22 x>c
Find c such that the function

Answer 78

@tkhunny

Answer 79

I would guess that the qeustion reads a little differently than you have presented it. We shoudl find a 'c' such that the mixed functino in COTINUOUS. No?

Can you graph both y = x^2 - 6 and y = 8x-22. Surely they intersect, somewhere.

Answer 80

it says
is continuous everywhere.
c=

Answer 81

Okay, then do it! Graph away!

Of course, you can just solve algebraically.

Answer 82

yay solving is cooler! so what is the first step

Answer 83

You are going to have to do better than that. Try x^2 - 6 = 8x - 22

Answer 84

x^2-8x+28=0

Answer 85

no sorry thats wrong

Answer 86

x^2-8x+16=0

Answer 87

There it is. Keep going.

Answer 88

how do you factoze beacuse 2*8=16 but 8-2= not 8.

Answer 89

You should recognize a perfect square when you see one. Try (-4)*(-4)

Answer 90

ohhhh makes sense!
(x+4)(x+4)

Answer 91

NOT　ITS (X-4)(X-4)!

Answer 92

hello i have to go but I will be back iat around 11:05!

Answer 93

Ta-da! it's x = 4. See if you get the same from both of your original equatins.

Answer 94

hello im back are you there

Answer 95

yes its same 10!

Answer 96

@AravindG im struggling with the second question

Answer 97

You're done. It's important to know when you're done. Simply write the definition to clean up the conclusion.