Question

how do i find the domain of the function: F(x)=square root of x^2-4/ x^2+x-6. I know im suppose to cancel out both x^2 but i got stuck on the next step.

Answer 1

hi fairy princess :)
is the function
\(F(x)=\dfrac{\sqrt{x^2-4}}{x^2+x-6}\)
?

Answer 2

hey and yes thats how the problem should look like, my bad

Answer 3

no problem :)
do you know how to get the domain ?

Answer 4

of any function..

Answer 5

eh....to be honest its confusing. could u give me a example of how to find the domain? maybe that would help

Answer 6

sure!
if f(x) = 1/(x+2)
the denominator cannot be = 0 !
so, x\(\ne-2\)
and -2 is excluded from the domain.

for f(x) = \(\sqrt{x+2}\)
the quantity under square root sign must be non-negative ! (so that x is real number)
so, \(x+2\ge 0 \implies x\ge -2\)


got these 2 examples ???

Answer 7

so that f(x) is real number ***

Answer 8

ohh...yeah i think i get it. like if its negative, it would make it undefined right?

Answer 9

yup, undefined or complex(to be precise)

so lets take your F(x)
the denominator cannot be 0, right ?
so can you solve for x^2+x-6=0 ??
quadratic equations?

Answer 10

yeah that way we'll find a answer to x

Answer 11

the 2 values you get for 'x' will be excluded from the domain,
so, can you find those 2 values of x ?

Answer 12

ohhh like x= positive or negatvie answer?

Answer 13

but i thought x^2 over the other x^2 would cancel both out each other.

Answer 14

not actually...
\(\Large \dfrac{x^2+stuff}{x^2+otherstuff}\ne \dfrac{stuff}{otherstuff}\)

Answer 15

oh okay thats where i was getting confused at

Answer 16

cool,
then what about x^2+x-6 = 0 ??
can u solve it ?

find 2 numbers whose sum is +1 and product = - 6 ???

Answer 17

wouldnt that be -7+1?

Answer 18

no,
i need sum to be 1 and product to be -6
try numbers like 2,3,-2,-3 ...

Answer 19

-3?

Answer 20

2 numbers ?

Answer 21

-2 and 3

Answer 22

correct!

Answer 23

so, x^2+x-6 = 0
will be
(x+3)(x-2) = 0
got this ?

Answer 24

yay! okay so basically its finding the sum of 1 and that'll help get negative 6. and yeah i got it

Answer 25

cool, so
(x+3)(x-2) = 0
gives x=-3, x=2
these values will be excluded from the domain, because they will make the denominator =0
got this part?

Answer 26

yeah and its cause the denominator cant be zero?

Answer 27

right! denominator = 0 is undefined

Answer 28

now for numerator, which is under square root
the quantity under square root must be non-negative, right?
so, solve for \(x^2-4 \ge 0\)
can u ?

Answer 29

right and yeah all u gotta do is add four then \[x ^{2}=4\]

Answer 30

then youll square root both and youll get postive or negative two right?

Answer 31

good!
so,
\(x\ge 2, \: x\le-2\)
any doubts ?

Answer 32

hmmmm no not really, i get it now much better. just my teacher didnt put all of that into two parts. like doing the denominator and the numerator

Answer 33

ok, and lastly we write everything together in the domain
\(domain \rightarrow (x\ge 2,x\le -2, x\ne 2, x\ne -3)\rightarrow (x> 2,x\le -2, x\ne -3)\)

Answer 34

and since you are new here,
\[ \begin{array}l\color{red}{\text{W}}\color{orange}{\text{E}}\color{#e6e600}{\text{L}}\color{green}{\text{C}}\color{blue}{\text{O}}\color{purple}{\text{M}}\color{purple}{\text{E}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{P}}\color{purple}{\text{E}}\color{red}{\text{N}}\color{orange}{\text{S}}\color{#e6e600}{\text{T}}\color{green}{\text{U}}\color{blue}{\text{D}}\color{purple}{\text{Y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{}}\end{array} \]
have fun learning :)

Answer 35

thank you for the help and yeah ill try to have fun learning xD Thanks for the welcoming also

Answer 36

most welcome ^_^