Question

Find the domain of these functions:

Answer 1

\[f(x)=\frac{ \sqrt{x^2-4} }{ x^2-9 }\]
\[g(x)=\ln(3x-2)\]

Answer 2

@agent0smith
@electrokid
@RadEn

Answer 3

for f(x) = sqrt(x^2-4)/(x^2-9)
so that the real values, x^2 - 4 >= 0
solve for x

also, the denominator cant be equal 0
it means x^2 - 9 not equal 0
solve for x

Answer 4

so \[x \neq 0,2\]

Answer 5

u just trying solve for the second case, right ?
let it is equal = 0
so, x^2 - 9 = 0
factor out!
(x+3)(x-3) = 0
the zeroes does satisfied for x = -3 or x = 3
so, x≠-3, 3

Answer 6

@abb0t

Answer 7

how about g(x)

Answer 8

The domain is all the values a variable can take on without causing an invalid function. So what values can you plug into "x" for g(x) that will make it invalid.

Answer 9

0

Answer 10

so domain is \[x \neq 0\]

Answer 11

@lilfayfay domain of the first function is not x≠0,2
reread RadEn's post:

so that the real values, x^2 - 4 >= 0
solve for x

also, the denominator cant be equal 0
it means x^2 - 9 not equal 0
solve for x

Answer 12

yeah i got the first one, i dont get the 2nd one

Answer 13

For g(x)=ln(3x−2)

3x-2>0

that will give the domain

Answer 14

so the domain is x>2/3

Answer 15

What was your domain for \[\Large f(x)=\frac{ \sqrt{x^2-4} }{ x^2-9 }\]

Answer 16

\[x \neq \pm 3\]

Answer 17

"so the domain is x>2/3" correct for g(x)

Answer 18

can you help me with another one please? ^^

Answer 19

We will not do your homework for you :3

Answer 20

x≠±3 is not correct for f(x)

\[\Large f(x)=\frac{ \sqrt{x^2-4} }{ x^2-9 }\]

reread this (both parts):

x^2 - 4 >= 0
solve for x

also, the denominator cant be equal 0
it means x^2 - 9 not equal 0
solve for x

Answer 21

but this is openstudy :(

Answer 22

and it would really help prepare me for the AP Calculus test next week

Answer 23

You need BOTH of those parts for the domain of f.