Question

Domain of sqrt(x^2+4)

Answer 1

\(\large\color{black}{ x^2+4 }\) is always a positive number.

Answer 2

The domain are the possible values for x that will produce a output y

Answer 3

Are you saying all real numbers?

Answer 4

Yes, good job!

Answer 5

\(\large\color{black}{ x^2\ge0 }\) ( for all real values of x.)
that means that you will never get any imaginary values in this case, (no matter what x is).

Answer 6

Could you explain why?

Answer 7

Hm..its not quite clicking

Answer 8

0^2
5^2
(-5)^2
(6)^2
(-2523523)^2

are all either positive or 0 right?

Answer 9

If a square root function has a negative number underneath the root, the result is an imaginary number, you do not want that

Answer 10

Yes freckles.

Answer 11

so adding a positive number to any of those
is still going to be positive

Answer 12

Correct Dan, elaborate more?

Answer 13

Well ok yes freckles, elaborate more please.

Answer 14

Are you all saying that when there is a positive number underneath a sq rt its always going to be all real numbers?

Answer 15

the domain is therefore all real numbers
since x^2+4 is always positive

Answer 16

so you want

x^2 + 4 >= 0

any x value squared , even if it is a negative number, will result in a positive value
For example,

(-2)^2 = +4

Answer 17

yes, sure;
For example if you had: \(\large\color{black}{ f(x)=\sqrt{x-4} }\)

then you would know that when \(\large\color{black}{ x<4 }\) then you are taking the square root of a negative number.
This is the only case where a square root can be undefined at x=c i.e. when the value of c (that you plug in for x, makes the inside of the square root a negative.

In your case, you have: \(\large\color{black}{ f(x)=\sqrt{x^2+4} }\) and there you will never encounter a negative inside the root.

Answer 18

\[\sqrt{x-4} \text{ we have \to worry \because x-4 can be negative }\]

Answer 19

lol @SolomonZelman we thought of the same function

Answer 20

yeah, that is weird and awesome!

Answer 21

Is it a rule that when you have imaginary in the sq rt its always going to be all real?
If you can't fully solve it out, it's all real?

Answer 22

the rule is that when you have a negative inside the square root (or any EVENth root) that it is then undefined.

Answer 23

Like if you had: (again using,) \(\large\color{black}{ f(x)=\sqrt{x-4} }\)
and plugged in 3 for x,

you would be getting \(\large\color{black}{ f(x)=\sqrt{-1} }\) and that is undefined for any real value.

Answer 24

ok ok.

Answer 25

one more question

Answer 26

I mean give me a real number a, so that: \(\large\color{black}{a \times a=-1 }\)

Answer 27

not a possibility, right?

Answer 28

sure, go ahead

Answer 29

I am doing compositions. g(f(x)) = sqrt(x+4)^2 =x+4 correct?
Then I am doing the domain of it.
I have to take the restrictions from both don't I? There was no restriction for g(x) it was all real
and for f(x) it was x is greater than equal to 4.
g(f(x)) is sqrt(x+4) so I got a domain of x is greater than equal to 4.

Answer 30

you \(\large\color{black}{ f(x)=\sqrt{(x+4)^2} }\) is same as \(\large\color{black}{ f(x)=x+4 }\)

Answer 31

and a polynomial is by definition continuous over \(\large\color{black}{ (- \infty,+\infty) }\).

Answer 32

yeah

Answer 33

wait what?!

Answer 34

you are correct, and are there any restrictions, or not>?

Answer 35

crap.

Answer 36

what?

Answer 37

lol

Answer 38

you don't get something?

Answer 39

hold on. Lemme check this work real fast.