## anonymous one year ago If f(x)=(x+1)^−1 and g(x)=x−2, what is the domain of f(x)÷g(x)?

1. campbell_st

well rewrite it as $f(x) = \frac{1}{x + 1}$ then if you divide by g(x) its $\frac{1}{(x + 1)} \div (x -2)~~~or ~~~\frac{1}{x +1} \times \frac{1}{x -2}$ so the domain is all real x, except there are 2 excepts.... you need to find them....

2. anonymous

(−∞,−1] and [2,∞)?

3. anonymous

@campbell_st

4. campbell_st

well you have the correct values for the exclusions.... but if you look at the domain you stated, you could have x = 0.... so I'd say all real x, except $x \neq -1 or 2$

5. anonymous

i have the answer choice "all values of x" , (∞,−1),(−1,2), and (2,∞) , (−∞,2) and (2,∞) , & (−∞,−1] and [2,∞)

6. campbell_st

so looking at the problem... things don't match my thinking... as x = 1 is aslo between the values x = -1 and x = 2 substitute x = 1 and you still get a numerical value... so I'll let someone else who is smarter help you.

7. anonymous

That's okay, I appreciate your help anyways

8. anonymous

How about "If f(x)=2x^2−x−6 and g(x)=x^2−4, what's f(x)÷g(x)?

9. campbell_st

here is a graph of the curves...

10. campbell_st

well for this question... its simple, start by factoring $\frac{2x^2 - x - 6}{x^2 - 4} = \frac{(2x + 3)(x - 2)}{(x +2)(x-2)}$ and hopefully you can finish it from here.

11. anonymous

Thanks!