Question

Determine where the following function is NOT continuous.

g(x) = (3x+7) / (x*cos(2x) + x)

Answer 1

Is there anything that x could equal to give you an indeterminate form?

Answer 2

x=0 would make it not continous

Answer 3

You're looking for any values of x that make the denominator = 0.

Answer 4

No, there is no indeterminate form?

Answer 5

Yes..there are values of x that make the denominator = 0. The function is NOT continuous at those points.

Answer 6

sorry @ninoootran you're right, I mean undefined

Answer 7

I'm not sure about that, but if it is a circle then would there be " +2PI n"?

Answer 8

When you look at any fraction, after simplifying it, you simply need to make the denominator zero for there to exist a singularity

Answer 9

I just start learning about Continuity. That is all the information is given.

Answer 10

That makes it undefined. So we get a hole or an asymptote

Answer 11

The cos is what thronw me off

Answer 12

is it cosx or xcosx?

Answer 13

If it is xcosx, does the cosine value matter?

Answer 14

well @ 0 cos = 1 so that's okay but the + x on the bottom is what makes it not continuous

Answer 15

it is g(x) = (3x +7)/ (xcos(2x) +x)

Answer 16

@faceless it's not cosx+x

Answer 17

I know

Answer 18

so nino, you want to find some x value that makes your fraction not have an answer in order to show the function is not continuous. The easiest way to do this is to find an x that makes the bottom =0. Any ideas?

Answer 19

This is Calculus, right?

Answer 20

Are what other standard of Math?

Answer 21

@31356 calc 1 or alg 2 I think

Answer 22

Oh, okay.

Answer 23

I have looked over my note, exmple like

f(x) = (10x + 2) / (x^2 - x - 6)
then we can factoer the bottom
(x-3)(x+2)=0
3 and -2 is where conti doesn;t exist

Answer 24

it is cal 1

Answer 25

Okay

Answer 26

Why in your example, does the function not exist at those points?

Answer 27

can we use the same strategy for this one?

Answer 28

because if you plug in those number. the bottom would be 0

Answer 29

not really. Yes you could factor something out, but it's not really necessary

Answer 30

and yes good, that was the main point I was trying to get accross

Answer 31

So make xcosx+x=0

Answer 32

but can we do the same thing to this problem?

Answer 33

like
xcos(2x)+x = 0?

Answer 34

^yes that

Answer 35

is this legal?

Answer 36

What would be illegal about it? You have not done anything

Answer 37

this what i got

x= [arccos(-1)]/2

Answer 38

...you are overcomplicating it

Answer 39

what does x need to equal to make xcosx=0?

Answer 40

also, that'd be wrong too. as a side note

Answer 41

SO you think I should leave the problem as xcos(2x)+x = 0?

Answer 42

yes. Tell me, ____ times anything = 0.

Answer 43

0

Answer 44

so x=?

Answer 45

0

Answer 46

That was all there was to it. So now your bottom would =?

Answer 47

0

Answer 48

so your function has become _____?

Answer 49

undefined

Answer 50

so the graph is _____?

Answer 51

BUt is there a way to find out if there are more function that is not continuous?

Answer 52

? I don't understand your question

Answer 53

or plug in numbers is the only way?

Answer 54

well, sight is the easiest way. If you have a fraction, simplify and make the bottom zero. or just break the rules for that fn. ie ln(-1)

Answer 55

or know where there is an asymptote ie tan(x)

Answer 56

that's all there is to it. You are overcomplicating it

Answer 57

My teacher, he normally don;t just want one answer but answers.

Answer 58

Maybe I am thinking too much

Answer 59

There is only one answer. Is there any other time the bottom of this fn would =0?

Answer 60

If cos is a circle, and make a full rotation, does that count?

Answer 61

no

Answer 62

like
x = 0 + 2Pi n

Answer 63

the most important thing you are missing is that the cosine is multiplied by x. So if x is anything other than 0,

Answer 64

you have xcosx+(something greater than zero)

Answer 65

the cosine is extraneous info in this case

Answer 66

if you really wanted you could factor out the x from each term and end up with a bottom of x(cos2x+1)....OOh wait I was wrong there is another answer

Answer 67

sorry I was rushing

Answer 68

let me correct this. So we already said that if x=0 the bottom will =0 right? so now if cos2x+1=0 the bottom will also =0

Answer 69

so what x yields cosx=-1?

Answer 70

Where did you get cos2x + 1 from?

Answer 71

I factored out the x.

Answer 72

then what happen to the x?

Answer 73

Do you know how to factor?

Answer 74

if x= 0 don't the cos 2x = to cos 2(0)?

Answer 75

yes

Answer 76

yes, but anything times zero = 0 still right. factoring the bottom you get, x(cos2x +1)

Answer 77

xcos(2x) + x
x(cos(2x) + 1)

Answer 78

then what is your next step?

Answer 79

Make one or the other equal zero

Answer 80

we can't make both be 0?

Answer 81

you dont need to

Answer 82

if the outside x is = 0 then the whole product is 0

Answer 83

right

Answer 84

the inside by the same philosophy

Answer 85

Thank you, I think i got it

Answer 86

good :)