# Thread: Finding the domain of g(f(x))

1. ## Finding the domain of g(f(x))

How would I go upon solving these:

1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1

2.
For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/h.
F(x)=3-9x^3

thank you!

2. ## clarification

Could you clarify what is meant by (f/g)(-1/4)?

3. Originally Posted by pmcoura
How would I go upon solving these:

1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1

2.
For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/H.
F(x)=3-9x^3

thank you!
Think of $\displaystyle (f/g)(-1/4)$ the same as $\displaystyle (\frac{f(-1/4)}{g(-1/4)})$

For #2 i'm not quite sure what its asking, but i'm sure someone will jump in and get it for you.

4. Originally Posted by ixo
Think of $\displaystyle (f/g)(-1/4)$ the same as $\displaystyle (\frac{f(-1/4)}{g(-1/4)})$
Oh.. so the first problem is just a simple function problem where you plug in the x-values to solve it?

5. Originally Posted by Catapult14
Oh.. so the first problem is just a simple function problem where you plug in the x-values to solve it?
I'm thinking that's what it's asking but I'm not sure myself. But if (f/g) is multiplied by (-1/4) shouldn't I just multiply the denominator by -4? (the denominator being g)

6. Originally Posted by ixo
Think of $\displaystyle (f/g)(-1/4)$ the same as $\displaystyle (\frac{f(-1/4)}{g(-1/4)})$
Well, according to ixo, you are not multiplying (f/g) by (-1/4), but are plugging in (-1/4) into the functions f(x)=X^2-2 and g(x)=4x+1:

$\displaystyle (\frac{f(-1/4)}{g(-1/4)})$ = $\displaystyle (\frac{(-1/4)^2-2}{4(-1/4)+1})$

Have I understood this correctly?

7. If f(x) = 2x and g(x) = 3x and h(x) = 4x
then f(g(x)) = 2( 3x ) and f(h(x)) = 2( 4x)
similarly g(f(x)) = 3( 2x ) and g(h(x)) = 3( 4x )
as well as
f(x)g(x) = (2x)(3x)

Whatever is inside the parenthesis is what you place as x for the equation. If f(x) = 2x and they ask you to evaluate f(2) then its just simply 2(2). Similarly if g(x) = 3x and they ask you to solve g(5) its just 3(5). If its (f$\displaystyle \cdot$g)(3) thats the same as f(3)g(3) or 2(3) x 3(3) , similarly if its (f/g)(9) its the same as f(9)/f(9) or 2(9) / 3(9).

Hope this helps

8. I realized the first problem was really simple and stupid for me to ask. It's not really that I had a hard time solving it it's just that the questions on my homework are badly written. But does anyone know what the second problem is trying to ask?

9. Originally Posted by pmcoura
How would I go upon solving these:

1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1

2.
For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/h.
F(x)=3-9x^3

thank you!
I think.. it may mean that you simplify for f(x) (lowercase) for (3-9x) :$\displaystyle \frac{f(x)+h(x)-3+9x^3}{h}$. But i'm not entirely sure. Why wouldn't they use a different variable instead of having f and F.. thats what i'm not understanding.