# Find the domain of a function

I am trying to find the domain of the following equation and I am not sure that I am right or not.

$$f(x) = \sqrt{\dfrac{(2x-4)(5+x)}{(x+3)}}$$

I know that the result in the square root must be positive or 0... and that $x=2$ and $x=-5$ for the numerator.

The square root will fail if I use $x = -3$ .. does it mean that the domain must be $\ge -3$?

Thanks.

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You have to deal with two separate failure modes. First, you cannot have a $0$ denominator, so $-3$ cannot be in the domain of the function. Secondly, $$\frac{(2x-4)(5+x)}{x+3}$$ must be non-negative, so you must solve the inequality $$\frac{(2x-4)(5+x)}{x+3}\ge 0\;.$$

The fraction is obviously positive when all three of the expressions $2x-4$, $5+x$, and $x+3$ are positive, but it’s also positive when exactly one of them is positive; in all other cases it is negative or zero. Now $2x-4>0$ when $x>2$, $5+x>0$ when $x>-5$, and $x+3>0$ when $x>-3$, so the relevant break points and intervals are as shown here:

---------|----------------------|-----------------------------------|-------
-5                     -3                                   2


When $x<-5$, none of the expressions is positive, so the fraction is negative; when $-5<x<-3$, only $5+x$ is positive, so the fraction is positive; when $-3<x<2$, $5+x$ and $x+3$ are positive, so the fraction is negative; and when $x>2$, all three are positive, so the fraction is positive. Thus, the fraction is positive on $$(-5,-3)\cup (2,\infty)\;.$$ It is zero at $x=2$ and $x=-5$; neither of these makes the denominator zero, so they also belong in the domain, which is therefore $$[-5,-3)\cup[2,\infty)\;.$$

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Awesome! :) Thanks ! – Jonathan Oct 19 '11 at 0:29

Notice that $2x-4= 2(x-2)\ge 0$ precisely if $x\ge2$, and in a similar way you need to figure out for which values of $x$ the other factors, $5+x$ and $x+3$, are positive, and for which values of $x$ they are negative. Then you have to think about multiplying and dividing positive and negative numbers, to figure out for which values of $x$ the whole fraction is positive, and for which values of $x$ it is negative. And finally, you need to exclude $-3$ from the domain.

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