anonymous
  • anonymous
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
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I know for a fact that choice b is wrong.
anonymous
  • anonymous
@dan815 @Math2400 @Camila1315 @notyourdroid
myininaya
  • myininaya
what does { } mean?

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myininaya
  • myininaya
for the two fractions you can find where they are not continuous by finding when the denominator is zero
anonymous
  • anonymous
parenthesis
myininaya
  • myininaya
so h(x)=x-1?
anonymous
  • anonymous
yes
myininaya
  • myininaya
h(x)=x-1 is just a line then
myininaya
  • myininaya
do you have any questions?
anonymous
  • anonymous
so my suspicion was right lol h(X)=x-1 is my answer? lol
myininaya
  • myininaya
I didn't say that
myininaya
  • myininaya
did you find when the bottoms were zero (if any values) for the first two?
anonymous
  • anonymous
there continuous but I put I and ii as my answer before and I was told it was wrong.
myininaya
  • 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
  • anonymous
x=1 for both
myininaya
  • myininaya
x^2-1=0 x^2=1 x=1 or x=-1 right? but how is 1^2+1 zero?
myininaya
  • myininaya
so f is discontinuous at x=1 and also x=-1 but what do you notice about x^2+1=0?
anonymous
  • anonymous
same as the first.
myininaya
  • myininaya
no (1)^2+1 is not zero (-1)^2+1 is not zero
myininaya
  • myininaya
both (1)^2+1 and (-1)^2+1 is 1+1 which is 2
myininaya
  • 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
  • anonymous
okay...
myininaya
  • 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
  • ganeshie8
{x} usually refers to the fractional part of x
anonymous
  • anonymous
+ x + = + - x - = +
myininaya
  • 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
  • myininaya
there is no real solution to x^2+1=0
myininaya
  • myininaya
therefore g=3/(x^2+1) has no discontinuities ok so h(x)=x-1 or h(x)={x-1} @magy33 ?
myininaya
  • myininaya
I'm asking since @ganeshie8 pointed out { } means fractional part
anonymous
  • 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
  • myininaya
If it is there, it does
ganeshie8
  • ganeshie8
http://mathworld.wolfram.com/FractionalPart.html
anonymous
  • anonymous
ok. I think then that from my understanding the answer is ii and iii
ganeshie8
  • ganeshie8
I'm inclining toward ii only
myininaya
  • myininaya
me to if h={x-1} and not x-1
ganeshie8
  • ganeshie8
the fractional function `h(x) = {x-1}` is discontinuous at every integer
anonymous
  • 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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