f(x) = { x-[[x+1]] / (x^2 - 3x + 2), if xor=1 Discuss the continuity at x=0, x=1, and x=4

Question

f(x) = { x-[[x+1]] / (x^2 - 3x + 2), if x<1
          { (6-3(sqrt of x)) / (x-4), if x >or=1

Discuss the continuity at x=0, x=1, and x=4

kc_kennylau · Answer

Just do substitution to see if they are defined :)

If the values are defined, it's continuous, otherwise, not.

anonymous · Answer

Could you show me your answer with solution? :-s

kc_kennylau · Answer

For example, when x=4, the value is undefined as the denominator equals to 0. Therefore the curve is discontinuous at x=4.

Try to do the rest yourself :)

anonymous · Answer

But when you substitute it in the numerator, you also get 0, meaning you end up with an indeterminate form (0/0). Are you allowed to conjugate when you do f(4)?

kc_kennylau · Answer

oh sorry then you need to do limit :)

kc_kennylau · Answer

Ok then do the limit for the 3 cases to see if the limits converge :)

kc_kennylau · Answer

For example at x=4 the limit is 3/4 which makes it continuous?
@kittiwitti1 can you help him :)

agent0smith · Answer

indeterminate form (0/0) doesn't suddenly make a function continuous. Undefined is undefined.

kittiwitti1 · Answer

@kc_kennylau I don't know if you've read my profile yet but it says I'm still in Pre-Calc...

agent0smith · Answer

y= sinx/x is not continuous when x=0, even though it's 0/0.

kc_kennylau · Answer

@kittiwitti1 oh lol sorry

kittiwitti1 · Answer

Haha it's not your fault ^^
I don't blame you :p

kc_kennylau · Answer

yay :P

agent0smith · Answer

At x=1, make sure each of the functions on the left and right are equal.

agent0smith · Answer

Oh you will have to eval. the limit going from the left, and check that equals the function on the right.

agent0smith · Answer

At x=4 it's just not continuous though.

anonymous · Answer

But at x=4 when you evaluate f(4) you get -9/12. Then when you get the limit of f(x) as x approaches 4 you also get -9/12 ... Right?

kc_kennylau · Answer

No, when you evaluate f(4) you get undefined.

anonymous · Answer

you get (0/0)

anonymous · Answer

Conjugate f(4) and you get -9/12

agent0smith · Answer

Why do you keep saying "conjugate"? That's not relevant here.

agent0smith · Answer

But at x=4 when you evaluate f(4) you get -9/12. Then when you get the limit of f(x) as x approaches 4 you also get -9/12 ... Right?


Being continuous means the function MUST be defined AT that point.

kc_kennylau · Answer

But f(4)=$\dfrac00$=undefined$\dots$

anonymous · Answer

0/0 is indeterminate?

anonymous · Answer

and when you get something indeterminate, you either factor or conjugate?

kc_kennylau · Answer

undefined is undefined don't try to make it defined lol

kc_kennylau · Answer

I think you may have confused it with limit where we continuing calculating upon meeting $\dfrac00$ or $\dfrac\infty\infty$

agent0smith · Answer

Undefined means the function is... undefined at that point. That means discontinuous.

anonymous · Answer

ohkayyyyy how about x=1?

agent0smith · Answer

And you're using conjugate very wrongly, i've no idea what you mean by that.

anonymous · Answer

well..by conjugate i mean multiply the entire thing by 6+3sqr of x / 6 + 3 sqr of x...the numerator ends up 36 - 9x

anonymous · Answer

oh wait, never mind. forget the x = 4, sorry. i copied the wrong number

agent0smith · Answer

At x=1, make sure each of the functions on the left and right are equal.

Oh you will have to eval. the limit going from the left, and check that equals the function on the right.