# range of x

• Jan 1st 2009, 12:29 AM
range of x
$\displaystyle \frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0$.

Find the range of values of x that satisfy the inequality .
• Jan 1st 2009, 01:45 AM
mr fantastic
Quote:

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

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.
• Jan 1st 2009, 08:09 PM
Soroban

Here is another (very primitive) approach . . .

Quote:

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

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

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

The four values divide the number line into five intervals:

. . $\displaystyle \begin{array}{ccccccccc} --- &*& --- &*& --- &*& --- &*& --- \\ & \text{-}2 & & \text{-}\tfrac{1}{2} && 1 && 3 \end{array}$

Test a value in each of the intervals.

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

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

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

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

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

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