Question

plz help fan medal and testimonial

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

Answer 1

I know for a fact that choice b is wrong.

Answer 2

@dan815 @Math2400 @Camila1315 @notyourdroid

Answer 3

what does { } mean?

Answer 4

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

Answer 5

parenthesis

Answer 6

so h(x)=x-1?

Answer 7

yes

Answer 8

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

Answer 9

do you have any questions?

Answer 10

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

Answer 11

I didn't say that

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

x=1 for both

Answer 16

x^2-1=0
x^2=1
x=1 or x=-1
right?


but how is 1^2+1 zero?

Answer 17

so f is discontinuous at x=1 and also x=-1


but what do you notice about x^2+1=0?

Answer 18

same as the first.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

okay...

Answer 23

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

Answer 24

{x} usually refers to the fractional part of x

Answer 25

+ x + = +
- x - = +

Answer 26

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

Answer 27

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

Answer 28

therefore g=3/(x^2+1) has no discontinuities


ok so h(x)=x-1 or h(x)={x-1} @magy33 ?

Answer 29

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

Answer 30

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

Answer 31

If it is there, it does

Answer 32

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

Answer 33

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

Answer 34

I'm inclining toward ii only

Answer 35

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

Answer 36

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

Answer 37

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