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which of the following are continuus for all real values of x?
I. f(x)= x^2+5/x^2-1
ii. g(X)= 3/x^2+1
iii. h(X)={x-1}
choices-
ii and iii only
I and ii only
I only
ii only

- anonymous

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- anonymous

I know for a fact that choice b is wrong.

- anonymous

@dan815 @Math2400 @Camila1315 @notyourdroid

- myininaya

what does { } mean?

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## More answers

- myininaya

for the two fractions you can find where they are not continuous by finding when the denominator is zero

- anonymous

parenthesis

- myininaya

so h(x)=x-1?

- anonymous

yes

- myininaya

h(x)=x-1 is just a line then

- myininaya

do you have any questions?

- anonymous

so my suspicion was right lol h(X)=x-1 is my answer? lol

- myininaya

I didn't say that

- myininaya

did you find when the bottoms were zero (if any values) for the first two?

- anonymous

there continuous but I put I and ii as my answer before and I was told it was wrong.

- myininaya

I assume the first fraction is:
\[f(x)=\frac{x^2+5}{x^2-1} \\ \text{ can you solve } x^2-1=0 \\ \text{ the second fraction is } g(x)=\frac{3}{x^2+1} \\ \text{ can you solve } x^2+1=0\]

- anonymous

x=1 for both

- myininaya

x^2-1=0
x^2=1
x=1 or x=-1
right?
but how is 1^2+1 zero?

- myininaya

so f is discontinuous at x=1 and also x=-1
but what do you notice about x^2+1=0?

- anonymous

same as the first.

- myininaya

no
(1)^2+1 is not zero
(-1)^2+1 is not zero

- myininaya

both (1)^2+1 and (-1)^2+1 is 1+1 which is 2

- myininaya

if I write x^2+1=0
as x^2=-1
If i try to think of a positive number to plug in...I'm gonna have to square it
but you know a positive times a positive =?
or
if i try to think of a negative number to plug in...I'm gonna have to square it
but you know a negative times a negative=?
or
if I plug in 0 which still isn't gonna work because 0^2 is not -1

- anonymous

okay...

- myininaya

I'm asking you to tell me what a positive times a positive will give
and
what a negative times a negative will give

- ganeshie8

{x} usually refers to the fractional part of x

- anonymous

+ x + = +
- x - = +

- myininaya

right x^2 is always positive or 0
-1 is negative
there is no way x^2 can even be -1 over the real numbers

- myininaya

there is no real solution to x^2+1=0

- myininaya

therefore g=3/(x^2+1) has no discontinuities
ok so h(x)=x-1 or h(x)={x-1} @magy33 ?

- myininaya

I'm asking since @ganeshie8 pointed out { } means fractional part

- anonymous

h(x)= x-1
but does the {} really have anything to do with the equation itself? I askto be sure for next time :)

- myininaya

If it is there, it does

- ganeshie8

http://mathworld.wolfram.com/FractionalPart.html

- anonymous

ok. I think then that from my understanding the answer is ii and iii

- ganeshie8

I'm inclining toward ii only

- myininaya

me to if h={x-1} and not x-1

- ganeshie8

the fractional function `h(x) = {x-1}` is discontinuous at every integer

- anonymous

lol I was half right. thnks for the help guys. lol
I still have more questions to go before im done lol I see if I can answer if not ill post lol

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