Find the solution:

great_math · #1 $Old$ 10-23-2008, 12:20 PM

Find the solution: $\frac{(3x-1)^3(x-2)(-5x-2)^2}{(x+2)^7} > 0$

Henderson · #2 $Old$ 10-23-2008, 03:58 PM

You have four critical points where things could change: $x = \frac{1}{3}, 2, -\frac{2}{5}, -2$ .

Plot those four points on a number line and make a sign diagram by plugging in a number from each of the 5 regions you've created to see if you get a positive or a negative.

Soroban · #3 $Old$ 10-23-2008, 09:08 PM

Hello, great_math!

Edit: Henderson has already supplied the analysis . . .

Quote:

Find the solution: . $\frac{(3x-1)^3(x-2)(-5x-2)^2}{(x+2)^7} \;>\; 0$

Consider graphing the function.

There are x-intercepts at: . $x \;=\;\tfrac{1}{3},\;2,\;\text{-}\tfrac{2}{5}$ (multiplicity 2)

There is a vertical asymptote: . $x \:=\:\text{-}2$
. . Note: there is a horizontal asymptote, $y = 0$

Testing values in each interval, we find:

. . $\begin{array}{cc}(\text{-}\infty,\:\text{-}2) & \text{negative} \\ \\[-4mm] \left(\text{-}2,\:\text{-}\tfrac{2}{5}\right) & \text{positive} \\ \\[-4mm] \left(\text{-}\tfrac{2}{5},\:\tfrac{1}{3}\right) & \text{positive} \\ \\[-4mm] \left(\tfrac{1}{3},\:2\right) & \text{negative} \\ \\[-4mm] (2,\:\infty) & \text{positive} \end{array}$

The graph looks like this:

Code:

              :         |
              :*        |                  *
              :         |                *   *
              : *       |*                      *
              :  *    * |  *            *           *
  - - - - - - : - - o - + - -o- - - - -o- - - - - - - - -
      *       :   -2/5  |   1/3*     * 2
          *   :         |         *
            * :         |
             *:         |
             -2         |

Solution: . $\left(\text{-}2,\,\text{-}\tfrac{2}{5}\right) \cup \left(\text{-}\tfrac{2}{5},\,\tfrac{1}{3}\right) \cup (2,\,\infty)$

o_O · #4 $Old$ 10-23-2008, 10:47 PM

As if you've made that with ASCII.