Question

precal help needed: see attachment

Answer 1

\[\large\rm f=\frac1x\qquad\qquad\qquad g=\sqrt{x+5}\]

f/g is just going to end up being division.\[\large\rm \frac{\color{orangered}{f}}{\color{royalblue}{g}}=\frac{\color{orangered}{\frac1x}}{\color{royalblue}{\sqrt{x+5}}}\]Our answer will have to include the domain restrictions applied to f,
as well as the domain restrictions applied to g,
and also g can no longer be zero because it's in the denominator.
So that provides another restriction.

Answer 2

Recall that in the land of math we cannot divide by 0.
So for our function f, x=0 is NOT included in the domain.

What about g?
Any ideas?
What values can we not plug into a square root?

Answer 3

about g: it can't be 0 because it's a denominator
we can't plug 0 into the square root?

Answer 4

can we eliminate x or no?

Answer 5

Well g has more restrictions than that.
We can't take the square root of a negative number, ya?\[\large\rm \sqrt{-4}=\text{no bueno}\]

Answer 6

But yes, the stuff under the root also can not be zero,\[\large\rm x+5\ne0\]But as I pointed out, it also must be positive, right?\[\large\rm x+5>0\]

Answer 7

ohh okay i'm following..

Answer 8

Subtracting 5 from each side,\[\large\rm x>-5\]This is the domain restriction being applied to g.
The strict inequality takes care of the possibility of the denominator being zero. We're excluding x=-5 with this `strict` inequality.

Answer 9

okay.. so now it becomes just -5?

Answer 10

What becomes -5? :o

Answer 11

So overall, we determined this:

~ x must be greater than -5
~ x can not be equal to 0.

All other x values are allowed.
Do you know how to write this in interval notation? :)

Answer 12

\[\frac{ 1/x }{ 5 }\]

Answer 13

\[-5 < x \neq 0\]

Answer 14

Determining the domain doesn't really `change` what the function looks like. This is more like.. work that we're doing on the side.

Answer 15

ohh okay

Answer 16

No no, I mean like the answer choices are written out.

For example, if I say:
~ x cannot equal 7

I would write the domain like this:
\(\large\rm (-\infty,7)\cup(7,\infty)\)
We're including every negative number up to 7,
and every positive number above 7.
We place `rounded brackets` around 7 to show that the value is excluded.

Answer 17

Ohhh okay, not sure how to do that for 5

Answer 18

Well, we determined that x must be greater than -5.
So in this case, we'll be completely excluding all of the negatives below that value.\[\large\rm (-5,\]This is how we'll start out, ya?
-5 is the lowest we can go.

Answer 19

We would like to write this,
\(\large\rm (-5,\infty)\)
But we can't.
We have to stop at 0 and exclude that value, yes?

Answer 20

Too confusing? :o

Answer 21

okay -5 is included because x must be greater than -5, so all values below it are not included.
One question, why the infinity sign?

Answer 22

-5 is excluded, hence the rounded bracket.
If -5 was included in our interval, it would have a square bracket to indicate that.
\(\large\rm [-5,\)

Answer 23

ohh okay, noted.

Answer 24

So let's look at the restrictions separately.
This is what's going on with our function g:
~ x must be greater than -5.

So our domain is... all numbers... greater than -5.
So that's from -5 all the way up to positive infinity.