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 
Newbie Joined: Nov 2011 Posts: 27 Thanks: 0  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 ... 
November 5th, 2011, 08:21 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Inequality Hello, jattrockz! Quote:
^3\,\,6x^2\,+\,5x\,+\,12" /> [color=beige]. . [/color]When is it positive? [color=beige]. . [/color]When is its graph above the xaxis? The function factors:[color=beige] .[/color]x\,+\,1)(x\,\,3)(x\,\,4)" /> [color=beige]. . [/color]Its xintercepts are:[color=beige] .[/color] The graph looks like this: Code:   *   * * 1 *  * *    o  +    o    o    *  3 * 4  *   1\:<\\:<\:3)\:\cup\x\:>\:4)" />  

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