1. ## domains of functions

f(x)=x^3-x^2+4 and g(x)=x^2-1
find:
-the domain of (f/g)(x)
-f(g(x))

thanks!!
also what does the x have to do with finding the answers when they tell you to find (f/g)(x) ??

2. Originally Posted by ss103
f(x)=x^3-x^2+4 and g(x)=x^2-1
find:
-the domain of (f/g)(x)
-f(g(x))

thanks!!
also what does the x have to do with finding the answers when they tell you to find (f/g)(x) ??
Hi

$\displaystyle \frac{f}{g}(x) = \frac{f(x)}{g(x)}$

$\displaystyle \frac{f}{g}$ is defined for all values of x for which $\displaystyle g(x) \neq 0$

$\displaystyle g(x) = x^2 - 1 = (x+1)(x-1)$ is equal to 0 for x=-1 or x=1

$\displaystyle f(g(x)) = f(x^2 - 1) = (x^2 - 1)^3-(x^2 - 1)^2+4$

$\displaystyle f(g(x)) = x^6 - 3x^4 + 3x^2 - 1 - x^4 +2x^2 - 1 + 4$

$\displaystyle f(g(x)) = x^6 - 4x^4 + 5x^2 + 2$

3. thank you! but what would the domain be in interval notation?

4. g(x) is equal to 0 for x=-1 or x=1

Therefore the domain of f/g is ]-oo,-1[ U ]-1,1[ U ]1,+oo[

5. Originally Posted by ss103
f(x)=x^3-x^2+4 and g(x)=x^2-1
find:
-the domain of (f/g)(x)
-f(g(x))

thanks!!
also what does the x have to do with finding the answers when they tell you to find (f/g)(x) ??

1) Think of as: $\displaystyle \frac{x^3-x^2+4}{x^2-1}$. So one can that $\displaystyle x\neq 1, -1.$

2) $\displaystyle f(g(x))$ no problem with any $\displaystyle x$ value. So the domain is $\displaystyle \mathbb{R}$.