# Thread: Find the domain of each rational function. Do not graph the function.

1. ## Find the domain of each rational function. Do not graph the function.

Hi, I need help on two problems.

23. f(x)=2x^2+7x-2/x^2-25 --> the answer for that one is : (- negative infinity, -5)U(5,5)U(5, positive infinity) My question is why is the answer (5,5) included in there.

25. f(x)= x-1/x^3-x --> the answer is (negative infinity,-1)U(-1,0)U(0,1)U(1,positive infinity) I had no idea how to do this problem.

All help is appreciated.

Thank you.

2. ## Re: Find the domain of each rational function. Do not graph the function.

Originally Posted by math951
Hi, I need help on two problems.

23. f(x)=(2x^2+7x-2)/(x^2-25) --> the answer for that one is : (- negative infinity, -5)U(5,5)U(5, positive infinity) My question is why is the answer (5,5) included in there.

25. f(x)= (x-1)/(x^3-x) --> the answer is (negative infinity,-1)U(-1,0)U(0,1)U(1,positive infinity)
in future, please use parentheses as grouping symbols

#22 it should be $(-5,5)$, not $(5,5)$

$\dfrac{2x^2+7x-2}{x^2-25} = \dfrac{2x^2+7x-2}{(x-5)(x+5)}$

the denominator, $x^2-25 = 0$ at $x = \pm 5$. the vertical asymptotes at these two x-values break up the real number line into three regions ...

$(-\infty,5) \cup (-5,5) \cup (5,\infty)$

#25 $\dfrac{x-1}{x^3-x} = \dfrac{x-1}{x(x^2-1)} = \dfrac{x-1}{x(x-1)(x+1)}$

vertical asymptotes at $x=0$ and $x=-1$ ... a point discontinuity (a "hole") at $x=1$

domain is $(-\infty,-1) \cup (-1,0) \cup (0,1) \cup (1, \infty)$

3. ## Re: Find the domain of each rational function. Do not graph the function.

Originally Posted by math951
Hi, I need help on two problems.

23. f(x)=2x^2+7x-2/x^2-25 --> the answer for that one is : (- negative infinity, -5)U(5,5)U(5, positive infinity) My question is why is the answer (5,5) included in there.

25. f(x)= x-1/x^3-x --> the answer is (negative infinity,-1)U(-1,0)U(0,1)U(1,positive infinity) I had no idea how to do this problem.

All help is appreciated.

Thank you.
23) the answer is wrong. The correct answer is $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$

Someone left out a negative sign.

25) $f(x) = \dfrac{x-1}{x^3-x} = \dfrac{x-1}{x(x^2-1)}$

the points where this is undefined are $x=0, -1, 1$ and thus the domain is

$(-\infty, -1)\cup (-1 ,0) \cup (0,1) \cup (1, \infty)$