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 November 5th, 2011, 07:31 PM #1 Member   Joined: Nov 2011 Posts: 37 Thanks: 0 Inequality Solve each inequality graphically. State the solution and graph the solution: a) (X^3) – (6x^2) +5x+12 >0 b) X/(x^2 -2x- >0
November 5th, 2011, 08:08 PM   #2
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Joined: Nov 2011

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Re: Inequality

I do not know right answer but in my head))) exist this:
X/(x^2 -2x- >0
for the (x^2 -2x- :
x = [ 2 +/- square_root( 2^2 - 4*(- ) ] / 2 = 1 +/- square_root( 4 + 32 )/2 =
= 1 +/- 6/2 = 1 +/- 3 = { 4, -2 } (x1 = 4 and x2 = -2)

then we can to write
x/(x^2 -2x- = x / [ (x - x1)*(x - x2) ] = x / [ (x - 4)*(x + 2) ]

I think this
x do not equal 4 and -2 because we will have division by zero (but we used this numbers to transform equation)

we have x from ( - infinity) to (+ infinity)
then

case ( x < - 2 ) : (-) /[ (-)*(-) ] < 0 /* because x < 0, x - 4 < 0 , x + 2 < 0 as conclusion from that :: x < - 2 */

case ( -2 < x < 0 ) : (-) /[ (-)*(+) ] > 0 /* because x < 0, x - 4 < 0 , x + 2 > 0 as conclusion from that :: -2 < x < 0 */

case ( 0 < x < 4 ) : (+) /[ (-)*(+) ] < 0 /* because x > 0, x - 4 < 0 , x + 2 > 0 as conclusion from that :: 0 < x < 4 */

case ( 4 < x ) : (+) /[ (+)*(+) ] > 0 /* because x > 0, x - 4 > 0 , x + 2 > 0 as conclusion from that :: 4 < x */

But ! You need to write it in proper form, which is demanded by teacher, book, because I do not know the rules of ...
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November 5th, 2011, 08:21 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Inequality

Hello, jatt-rockz!

Quote:
 Solve each inequality graphically. State the solution and graph the solution: [color=beige]. . [/color]$a)\;x^3\,-\,6x^2\,+\,5x\,+\,12\:>\:0$

$\text{W\!e have a cubic function: }\:y \:=\^3\,-\,6x^2\,+\,5x\,+\,12" />
[color=beige]. . [/color]When is it positive?
[color=beige]. . [/color]When is its graph above the x-axis?

The function factors:[color=beige] .[/color]$f(x) \:=\x\,+\,1)(x\,-\,3)(x\,-\,4)" />
[color=beige]. . [/color]Its x-intercepts are:[color=beige] .[/color]$-1,\:3,\:4$

The graph looks like this:

Code:
            |
|                   *
|
| *                *
-1 * |     *           *
- - - o - + - - - o - - - o - - -
*    |       3   *   4
|
*     |
|

$\text{Solution: }\-1\:<\\:<\:3)\:\cup\x\:>\:4)" />

$\begin{array}{cccccccccc}\text{Graph: }=&--=&o=&o=&--=&o=&-1=&3=&4 \end{array}=$

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# -2x2 5x 12<=0 in equality

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