Results 1 to 13 of 13

Math Help - Inequality Help

  1. #1
    Banned
    Joined
    Oct 2007
    Posts
    158

    Inequality Help

    Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0 From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by OzzMan View Post
    Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0 From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.
    There are various and sundry ways to look at it but I think the most visual, and the easiest to explain is the graphing method.

    Graph (x + 1)(x - 2)(x - 3) (see below).

    Your solution set is where the graph dips below the x axis, x \in (-\infty, -1) \cup (2, 3)

    -Dan
    Attached Thumbnails Attached Thumbnails Inequality Help-cubic.jpg  
    Last edited by topsquark; February 14th 2008 at 04:31 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2007
    Posts
    158
    Wow. Yea thanks man.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2007
    Posts
    158
    Oh is my solution set always the part of the graph that dips below the x axis?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by OzzMan View Post
    Oh is my solution set always the part of the graph that dips below the x axis?
    Only because you have a "<" in the problem. You need to change how you get your answer according to whether you have a " \leq", " \geq", "<", or ">" in the problem.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2007
    Posts
    158
    Could you help me with this one? Producing the graph and such and telling me what to look for in the graph. \frac {x-8}{3-x}<0 The graph is confusing me.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by OzzMan View Post
    Could you help me with this one? Producing the graph and such and telling me what to look for in the graph. \frac {x-8}{3-x}<0 The graph is confusing me.
    Well, once again you have a "<" so you are looking for all points on the graph below the x-axis. (I never said it, but when setting up a problem like this you always want to get the RHS to be 0.)

    So the only trick is to see where
    \frac{x-8}{3-x} = 0

    x - 8 = 0 for x \neq 3

    x = 8.

    So the solution set will be x \in (-\infty, 3) \cup (8, \infty).
    (The hyperbola moves from negative to positive over the vertical asymptote x = 3, if that's what's bothering you.)

    -Dan
    Attached Thumbnails Attached Thumbnails Inequality Help-hyperbola.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2007
    Posts
    158
    Alright. So the answer is T>8\quad T<3 right?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by OzzMan View Post
    Alright. So the answer is T>8\quad T<3 right?
    Yes, though I'm not sure where you got the T from.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Oct 2007
    Posts
    158
    Oh yeah T was the variable used in my book instead of X. It is still a little confusing though, sadly. I don't know why but from the looks of the graph it looks like the opposite of what I listed the answers as above. But I'm going to listen to you anyways.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by OzzMan View Post
    Oh yeah T was the variable used in my book instead of X. It is still a little confusing though, sadly. I don't know why but from the looks of the graph it looks like the opposite of what I listed the answers as above. But I'm going to listen to you anyways.
    Make sure you understand it. You are looking for regions on the graph where the graph is below the x-axis. That is how you determine what intervals are part of the solution set.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Banned
    Joined
    Oct 2007
    Posts
    158
    Yea i think i see what you are saying though about the hyperbola moving from negative to positive over the vertical asymptote of x=3. Thanks again.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jul 2006
    Posts
    11

    number line method solution

    Quote Originally Posted by OzzMan View Post
    Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0 From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.
    (x+1)(x-2)(x-3)< 0
    The solution range are -1, 2, 3 by using number line method.
    ________-____+_______-___
    | | | | | | | | | | | | | | | |
    -2-1 01 2 3
    Since
    (x +1)(x-2)(x-3)=0
    Substitute -2 into x in quadrate equation.
    (-2 +1)(-2-2)(-2-3) = -1* -4* -5 = -20
    x< -1, -1< x< 3, x >3.
    The inequality solution is x < -1 and x > 3


    from clement
    cokhale@yahoo.com
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 09:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 02:16 PM
  3. inequality
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: July 25th 2010, 07:11 PM
  4. Inequality help
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 8th 2010, 07:24 AM
  5. Inequality :\
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 12th 2009, 02:57 PM

Search Tags


/mathhelpforum @mathhelpforum