Question

domain of y=x/(x^2-9)

Answer 1

the domain is a set of numbers.

When you get functions like this also pay attention to the denominator.

Answer this question: what value for x will give you '9'

Answer 2

\[y=\frac{ x }{ x^2-9 }\]

Answer 3

9

Answer 4

...look at the denominator. For what value of x (if you substitute in a number) will give you 9 where it says x^2?

Why am i asking you this?

Because the function is U N D E F I N E D if the DENOMINATOR = 0.

So with problems like this always solve for the denomintor = 0 and exclude that number(s) from the domain that makes the denominator = to 0. And in this case it will be THREE. Because 3^2 = 9. And 9-9 = 0 !!!!!!!!!!!!!

THUS

Your domain is

\[x \epsilon (-infinity,3)U(3,+infinity)\]

Answer 5

I gave you the domain in INTERVAL NOTATION. Do you understand interval notation as I have written?

Answer 6

yes

Answer 7

cant u also right x cannot equal 3 or -3

Answer 8

You were wrong when I asked you your question of what value of x will be 9 in the DENOMINATOR (Remember, we don't care about the numerator. Because it will be valid for any number of x.)

Simply rewite the denominator in cases like this and solve for 0. That x value that makes the denominator 0 (in this case 3) is the value that x CANNOT BE. WHY!??? BECAUSE it will make the function U N D E F I N E D ------->

If you divid any number by 0 you get an ERROR / UNDEFINED.

you understand?

Answer 9

Yes, you can also write it like

x=/= 3 or x=/=-3.

Answer 10

I know I was wrong im asking can I also write x cannot equal -3 or 3 instead of wat u wrote? yes or no question

Answer 11

ok thank u

Answer 12

now I took the first derivative and need to find critical numbers

Answer 13

i set x^2-9=0 and i got -3 and +3 but since there not in the domain i don't have critical numbers right?

Answer 14

have you solved for the FIRST DERIVATIVE of the whole function yet?

Answer 15

yes

Answer 16

then i set x^2-9=0 and i got -3 and +3 but since there not in the domain i don't have critical numbers right?

Answer 17

thats not correct.

Answer 18

i got 1/x^2-9 for first derivative

Answer 19

post dy/dx (the derivative function) for me please.

Answer 20

Oh, is THAT the first derivative?

Answer 21

yes that's the answer to my first derivative

Answer 22

\[dy/dx(x/x^2-9) = 1/x2-9\]

Answer 23

ok, what you need to do is the following:

1.) Solve for f'(x) = 0 or dy/dx = 0)

Do you know how to do that?

Answer 24

i did

Answer 25

no, you solved for the denominator = 0. Thats for the DOMAIN.

Answer 26

c=+-3

Answer 27

oh

Answer 28

so it would just be 1=0

Answer 29

You need to solve for the DERIVATIVE FUNCTION = 0.

CRITICAL POINTS are those points in which the GRADIENT (Derivative) of a function is =0.

Answer 30

0=1/x^2-9

Answer 31

so there are no critical numbers right

Answer 32

hello?

Answer 33

i have another question too

Answer 34

hello?

Answer 35

y=x/(x^2-9)
for y to be defined x^2-9\[\neq 0\],x^2\[\neq9\]
x≠3
x<-3 or x>3
combining
domain is (-∞,-3)∪(3,∞)

Answer 36

thank u :)

Answer 37

the critical points are -3 and 3, is that what you got?