# How do I solve inequalities of the form $\frac{ax}{b}\geq0$?

I need to find the maximum domain for $f(x) = \sqrt{\frac{4x+13}{(x+5)(2-x)}}$

Therefore, I should solve the inequality

$$\frac{4x+13}{(x+5)(2-x)} \ge 0$$

I don't remember how to solve inequalities for the form

$$\frac{ax}{b}\geq0$$

Normally, I move the $b$ to multiply to the right side, but, you know, since there is a zero, It'll just eat the $b$. Then the $a$. So I end up with $x\geq0$, which doesn't make sense.

All examples I find are not of such form, so I'm a bit lost.

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Are you looking for the domain of the function? – Mhenni Benghorbal Dec 7 '12 at 5:14
@MhenniBenghorbal: Yeah, I'm not sure why was I calling it maximum domain... – Zol Tun Kul Dec 7 '12 at 5:14
Note that $AB/C\ge0$ if and only if either all three of $A,B,C$ are at least zero, or exactly one of them is. – Gerry Myerson Dec 7 '12 at 5:15

My favorite way to solve these problems is using a sign chart. This is basically a number line that has the key information on it - the places where the quantity is positive, negative, zero and undefined. To make one, find the places where it equals zero first (remember, if a fraction is zero, the top must be zero and the bottom not zero). Then, find the places where it's undefined (where the bottom is zero). Then, "test" points in between your zeros and undefined points to check the sign. Here what the sign chart would look like for the expression

$$y=\frac{x+1}{(x-1)(x-2)}:$$

You should be able to replicate this for your situation! Don't forget that when you take the square root, the expression inside the root needs to be defined and $\geq 0$. So watch out for divide-by-zero.

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Multiplying or dividing doesn't work directly since you don't know if what you call the $a,b$ terms are positive or negative. Remember that multiplying/dividing an inequality by a negative sign switches the sign.

Here is the standard approach:

Find the sign of $4x+13$. Find the sign of $x+5$. Find the sign of $2-x$.

Once you have the sign of each term you can find the sign of the fraction. Looke where is positive....

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A related problem. We want

$$\frac{ g(x) }{ h(x)k(x) } > 0 \implies \left\{g(x)<0 \cap h(x)k(x)<0 \right\} \cup\left\{g(x)>0 \cap h(x)k(x)>0 \right\}.$$

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Yes...but...you've written $A\cup A$. – Gerry Myerson Dec 7 '12 at 5:56