Question

can someone please solve these for me :(
1. y = 2x^2 + 5 over x^2
2. y= 3x - 3 over x^2 - 1
3. y = x^2 + 5x+ 6 over x^2 + 6x + 9
4. y =x ^3 -8 over x^2 -8
5. y= 1 over 2x^2 + 3x - 7
please help me!

Answer 1

what does it mean to "solve"?

Answer 2

the question says " find any points of discontinuity for each rational function"

Answer 3

@amistre64

Answer 4

discontinuity in this case usually defines a value for which the function becomes undefined at. The values for when the denominators go to zero

Answer 5

oh okay , so could u plz answer them for me

Answer 6

answer them for you is a different website. We promote studying and learning of the material over handing out specifics.

I would stay and see if i could guide you thru them, but i have class starting pretty soon

Answer 7

3 and 5 seem to be about the trickest ones to solve for; the others are pretty straight forward

Answer 8

oh okay , so could u jus help me with the first problem ? like how do i solve it ?

Answer 9

y = 2x^2 + 5 over x^2

when does x^2 = 0 ?

Answer 10

y= 3x - 3 over x^2 - 1

when does x^2 - 1 = 0 ?

Answer 11

um .... i dont know

Answer 12

\[\frac{3x-3}{x^2-1}\]
When
\[{x^2-1}=0\], the function breaks and there is a discontinuity (you can't divide by 0)>
\[x^2-1=0\]
\[x^2=1\]
Do you know what to do from here?

Answer 13

no i dont know any of this at all

Answer 14

Are you familiar with the fact that you can't divide by 0?

Answer 15

yea i know that fact

Answer 16

A discontinuity is where the function breaks down (that is, the input doesn't make sense when you perform the function on it). For example, the function 1/x breaks when you insert x=0, as you can't divide by 0. Same thing here.

Answer 17

oh okay
so what would be the answer for the first question ?

Answer 18

Which x do you put into the function that makes there be a 0 in the denominator?

Answer 19

um ... 2x?

Answer 20

What number?

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