Quick Math Problem?

Question

How would I find the domain of these functions?

g(x) =

2
/
x²+x-12


and...


g(x) =

1
/
√x+2


Thank you so much.

Answer 1

these are called asymptotes, which means that the graph gets infinitely close to these values but never reaches them.

1st problem:
the asymptotes in this problem are the numbers which make the denominator = 0, because 2/0 is "undefined"

to solve you just set the denominator = to 0

x^2+x-12=0
(x+4)(x-3)=0
x=-4 and 3
since x can't be these values, the domain is written as
x<-4 or x>3 or -4
2nd problem:
This problem is a little different because there is nothing that will make the denominator = 0.
What would make the denominator impossible would be a negative number, because the squareroot of a negative is imaginary.
Any positive value for x would work though.

so the domain is x>=0 (x greater than or equal to 0)

Answer 2

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Answer 3

set the denominators not equal to zero
ex. X^2 +X -12=0, find out what that equals and that's what the domain cannot equal so it would be {X cannot equal -4, 3}

and the other one would be {X cannot equal 4}

Answer 4

You can't divide by zero.
x^2+x-12 = (x+4)(x-3) so domain of first one is all real numbers except -4 and 3.

Can't take square root of a negative number. so domain of the second one is all real numbers greater than or equal to zero.

Answer 5

The answer of CS is perfect but for the second problem the domain will be x>-2

Answer 6

I answered this one when you asked it before... look down

Answer 7

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