let f(x)=1/x and g(x)=x^2-9x
What two numbers are not in the domain of (f o g)

Question

let f(x)=1/x and g(x)=x^2-9x
What two numbers are not in the domain of (f o g)

astrophysics · Answer

Lets start off by setting it up as f(g(x)) first, go ahead and do that

decarr432 · Answer

1/(x^2-9x) right?

astrophysics · Answer

Perfect, $$f(g(x)) = \frac{ 1 }{ x^2-9x } = \frac{ 1 }{ x(x-9) }$$ I simplified it a bit more for you so it's easier to see (by factoring out an x).

astrophysics · Answer

Good so far?

decarr432 · Answer

Yeah

astrophysics · Answer

Ok now we can see the restrictions by $$x^2-9x \neq 0 $$ or you can see it clearly in the simplified form by setting $$x \neq 0 ~~~\text{and}~~~x -9 \neq 0$$ I sort of gave away one of the numbers not allowed haha but it's ok long as you understand it.

astrophysics · Answer

Again we solve for x

astrophysics · Answer

The reason I have set it cannot equal to 0 is because if we divide over 0 we get undefined right?

decarr432 · Answer

Yeah makes sense

decarr432 · Answer

greater than zero right?

decarr432 · Answer

Could it be a negative number?

astrophysics · Answer

Not quite, we're not working with square roots, lets think about this. So we have one of our restrictions as I showed you which was $$x \neq 0$$ now lets see what happens if we plug in the 0 in our function. $$f(g(0)) = \frac{ 1 }{ 0(x-9) } = \frac{ 1 }{ 0 }$$ uh oh undefined..that's not going to work, so it will be one of our two numbers not in the domain.

astrophysics · Answer

For the second number $$x-9 \neq 0$$ solve for x

decarr432 · Answer

x can be any number as long as the outcome does not equal zero correct?

astrophysics · Answer

Right

decarr432 · Answer

Okay so how do we figure out two separate numbers that are not in the domain?

decarr432 · Answer

one would be 9 right?

astrophysics · Answer

Right and the other is 0 we got that from (quoting myself lol)


 $\color{blue}{\text{Originally Posted by}}$ 
@Astrophysics
Ok now we can see the restrictions by $$x^2-9x \neq 0 $$ or you can see it clearly in the simplified form by setting $$x \neq 0 ~~~\text{and}~~~x -9 \neq 0$$ I sort of gave away one of the numbers not allowed haha but it's ok long as you understand it.
$\color{blue}{\text{End of Quote}}$

astrophysics · Answer

We looked at our function and saw our denominator and set that equal to 0

decarr432 · Answer

Okay so recap

decarr432 · Answer

Any number is allowed as long as it doesn't come out to be zero in the end?

astrophysics · Answer

0 in the `denominator`

astrophysics · Answer

because we get undefined

decarr432 · Answer

yeah but im scared to submit my answer lol

astrophysics · Answer

Why? We know our two numbers that are not in the domain, I suggest you go over everything in this post again

decarr432 · Answer

because it says g(x) x^2-9x

astrophysics · Answer

That's ok

decarr432 · Answer

its the square thats confusing me

decarr432 · Answer

Well I manned up and it was correct thanks bud again lol I owe you alot haha

astrophysics · Answer

No worries, just go over it again, don't let math scare you ;)! They will try to trick you but it's the same concept being applied over and over again

decarr432 · Answer

Okay last one I'll submit the question