Question

g(x) = √x2-3, find domain and range

Answer 1

is the sqrt include all?

Answer 2

yes

Answer 3

is that x^2 or 2x?

Answer 4

x^2-3 >= 0
x^2 >= 3
x>= sqrt 3
for domain^

Answer 5

\[x ^{2}\]

Answer 6

then?

Answer 7

well and x <= -sqrt(3)

Answer 8

why negative?

Answer 9

range is 0 to infinity

Answer 10

because -sqrt(3) or less squared is more than 3 thus never negative under the radicant

Answer 11

http://www.wolframalpha.com/input/?i=graph+sqrt%28x%5E2-3%29
ignore the red (complex) line

Answer 12

okay, thanks :)

Answer 13

the negative part comes from the line where x^2 = 3, so (+ or -) x>= sqrt3 then we have two equations x>= sqrt(3) and then we change the sign when we multiply the RHS to get the negative that we must retain so x <= -sqrt(3)

Answer 14

Domain: What x is allowed to be
Range: What y is allowed to be (Here y is renamed as g(x) )
I will assume you are working with the real numbers

You have \[g(x)=\sqrt(x^2-3)\]
But the inside of your radical must be greater than or equal to 0.
In other words the inside of the radical is 0 or positive.

In other words, you do not want to run into the problem that you will take the square root of a negative number (***because you are working with the real numbers)

So for the domain you have to solve:
\[x^2-3\ge0\] this is the inside of the radical, which says it's 0 or positive
\[x^2\ge3\] Added 3 to both sides
\[x \ge \sqrt3\] Took the square root of both sides
So the domain is exactly this last inequality. That is what x is allowed to be.

For range you simply start plugging in x's and see what you get for g(x) [or y].
when x=3 you get g(x)=0
And when you plug in bigger x's you find out g(x) simply increases, or gets bigger. So your range is \[g(x) \ge 0\]

Answer 15

the last part is not right the domain will include anything smaller and -sqrt(3)

Answer 16

You are correct zzr0ck3r.
I forgot to include in the domain something that is very important.
So from the equation
\[x^2 \ge 3\]
We will get \[x \ge \sqrt3\] or \[x \le -\sqrt3\]

@AmiraZainal
to understand this a bit easier let's try using nicer numbers:
if we had ended up with \[x^2 \ge 9\]
Your answer for x is which number when I multiply it by itself gives me at least 9?
And you will realize that when x=3 you get at least 9 since 3*3=9, so you get
\[x \ge 3\]
But if you try -3 * -3 you will also get at least 9
So your answer also includes
\[x \le -3\]