# How can I find the domain of this function?

• Oct 29th 2013, 04:27 PM
letranger9000
How can I find the domain of this function?
f(x) = (√1-2x)/(x^2 -9) + 3/(x^2 - 6x +8)

I thought:

x^2-9 -> (x-3)(x+3)

x^2-6x+8 -> (x-4)(x-2)

(√1-2x) (x-4)(x-2) + 3(x-3)(x+3)
---------------------------------------
(x-3)(x+3)(x-4)(x-2)

I just don't know what to do next and it just feels not right (Angry), please someone tell me how can I find the domain.
• Oct 29th 2013, 06:57 PM
Soroban
Re: How can I find the domain of this function?
Hello, letranger9000!

Quote:

Find the domain of: . $f(x) \;=\;\frac{\sqrt{1-2x}}{x^2 -9} + \frac{3}{x^2 - 6x +8}$

I thought:

$x^2-9 \:=\:(x-3)(x+3)$

$x^2-6x+8 \:=\:(x-4)(x-2)$ . Good!

The denominators must not be zero.
. . Hence: . $x \:\ne\:-3, 2, 3, 4$

From that radical: . $1-2x \:\ge\:0 \quad\Rightarrow\quad -2x \:\ge\:-1$
. . Hence: . $x \:\le\:\tfrac{1}{2}$

We want all numbers less than or equal to $\tfrac{1}{2}$, except $-3.$

Domain: . $(\text{-}\infty,\,\text{-}3) \,\cup\,\left(\text{-}3,\,\tfrac{1}{2}\right]$