Question

let f(x)=1/x and g(x)=x^2+4x what are two numbers NOT in the domain

Answer 1

So we know the first one is 0 because the same rule applies as last question right?

Answer 2

Well lets go over it step by step, I will let you try it yourself now and I will tell you if you're on the right path or not :)

Answer 3

Ok first step, what do we do?

Answer 4

f(g(x)) right :)

Answer 5

I got the answer but I want to work it out

Answer 6

Good, so how does it look

Answer 7

f(g(x))= 1/(x^2+4x) or 1/(x(x+4) right?

Answer 8

double )) on the last one

Answer 9

\[f(g(x)) = \frac{ 1 }{ x^2+4x } = \frac{ 1 }{ x(x+4) }\] looks good!

Answer 10

Keep going

Answer 11

0 is ruled out don't know how to write it how you do

Answer 12

because it would be undefined

Answer 13

Notice \[f(g(x)) = \frac{ 1 }{ \color{red}{x^2+4x} } = \frac{ 1 }{ \color{red}{x} \color{blue}{(x+4)} }\] then we set \[\color{red}{x \neq 0} ~~~~\text{and}~~~\color{blue}{x+4\neq0}\]

Answer 14

maybe I should make the x green haha in any case I think you get the point

Answer 15

Yeah undefined then we check for the second number with can only be one that makes zero when added by 4

Answer 16

The second one would have to be -4 because -4+4=0 so undefined

Answer 17

Yes

Answer 18

But you should also practice your algebra, as that is a very important tool

Answer 19

Okay teach me ol wise one how would I set this up

Answer 20

Plugging in a number is a good way to check

Answer 21

okay so like

Answer 22

That one would work right?

Answer 23

I am not sure what that is, \[\frac{ 1 }{ 1(1+4) }?\]

Answer 24

I was using 1 in place of x should I not?

Answer 25

why

Answer 26

I'm not sure what you're doing

Answer 27

I just choose the first number i thought of lol

Answer 28

I was finding numbers that worked with the problem and didn't give me an undefined number

Answer 29

Oh, we don't need to do that, all real numbers work except for 0 and -4 which we got from solving \[x \neq 0 ~~~ \text{and}~~~x + 4 \neq 0\]

Answer 30

Okay

Answer 31

The reason these problems exist is, to show you `what numbers we cannot` have in the domain, but all the other real numbers are allowed

Answer 32

I was saying plugging in the restrictions are a good way to check if your restrictions