range of x

**mathaddict** · January 1st 2009, 12:29 AM

$\frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0$ .

Find the range of values of x that satisfy the inequality .

**mr fantastic** · January 1st 2009, 01:45 AM

Originally Posted by mathaddict

$\frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0$ .

Find the range of values of x that satisfy the inequality .

Case 1: Solve (x + 2)(x + 1/2) < 0 and (x - 1)(x - 3) > 0 simultaneously.

Case 2: Solve (x + 2)(x + 1/2) > 0 and (x - 1)(x - 3) < 0 simultaneously.

**Soroban** · January 1st 2009, 08:09 PM

Hello, mathaddict!

Here is another (very primitive) approach . . .

$\frac{(x+2)(x+\frac{1}{2})}{(x-1)(x-3)} \:<\:0$ .

Find the range of values of $x$ that satisfy the inequality .

We see that the "critical values" are: . $\text{-}2,\:\text{-}\tfrac{1}{2},\:1,\:3$

The four values divide the number line into five intervals:

. . $\begin{array}{ccccccccc}<br /> --- &*& --- &*& --- &*& --- &*& --- \\<br /> & \text{-}2 & & \text{-}\tfrac{1}{2} && 1 && 3 \end{array}$

Test a value in each of the intervals.

On $(\text{-}\infty,\text{-}2)\!:\;\;f(\text{-}3) \:=\:\frac{(\text{-}1)(\text{-}\frac{5}{2})}{(\text{-}4)(\text{-}6)} \:=\:+\quad\hdots \text{ no}$

On $(\text{-}2,\text{-}\tfrac{1}{2})\!:\;\;f(\text{-}1) \:=\:\frac{(1)(\text{-}\frac{1}{2})}{(\text{-}2)(\text{-}4)} \:=\:{\color{red}-}\quad\hdots \text{ yes!}$

On $(\text{-}\tfrac{1}{2}, 3)\!:\;\;f(0) \:=\:\frac{(2)(\frac{1}{2})}{(\text{-}1)(\text{-}3)} \:=\:+\quad\hdots \text{ no}$

On $(1,3)\!:\;\;f(2) \:=\:\frac{(4)(\frac{5}{2})}{(1)(\text{-}1)} \:=\:{\color{red}-} \quad\hdots \text{ yes!}$

On $(3,\infty)\!:\;\;f(4) \:=\: \frac{(6)(\frac{9}{2})}{(3)(1)} \:=\: + \quad\hdots \text{ no}$

Therefore: . $(\text{-}2, \text{-}\tfrac{1}{2}) \:\cup\: (1,3)$

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