Results 1 to 3 of 3

Thread: range of x

  1. #1
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1

    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 .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by mathaddict View Post
    $\displaystyle \frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0$.

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, mathaddict!

    Here is another (very primitive) approach . . .


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

    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) $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Range of b
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jun 18th 2011, 08:26 AM
  2. range
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 10th 2011, 10:06 AM
  3. range
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Mar 27th 2010, 05:46 AM
  4. Replies: 6
    Last Post: Sep 16th 2009, 06:25 AM
  5. range, inter-quartile range
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Jun 19th 2006, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum