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Thread: Inequation with Maxima ,define all x

  1. #1
    Newbie MysteriousPP's Avatar
    Dec 2010

    Inequation with Maxima ,define all x

    OMG, I hope I post it in the correct forum, I don't want to get infracted

    Heyah, I have this problem to solve, I tried a bit and came to the point where I don't know how to continue. I think I just have to prove my results.. Dunno. I just want to add,that english is not my mother tongue, and I might name some things incorrectly, so just correct my mistake but don't friggin kill me :P . Here it is:

    Define all $\displaystyle x\in \mathbb{R}$ with $\displaystyle x\neq 1,2$ and:

    $\displaystyle \frac {(x^{16}+5)max\left \{ x^{2}+x-2,2x+4 \right \}}{(x-2)^{2}(x-1)}\geq \frac {(4x+8)(x^{16}+5)}{(x-2)(x^{2}-4)}$

    First I have to simplyfy , so I checked this first :

    $\displaystyle (x^{16}+5)\geq 1\geq 0$

    so I don't have to change the inequality sign:

    $\displaystyle \frac {max...}{(x-2)(x-2)(x-1)}\geq \frac {(4x+8)}{(x-2)(x^{2}-4)}$

    $\displaystyle (x-2)< 0 \Rightarrow \frac {max...}{(x-2)(x-1)}\leq \frac {(4x+8)}{(x^{2}-4)}$

    Here I've changed the inequality sign and I also noticed that:

    $\displaystyle x^{2}-4=(x-2)(x+2)\Rightarrow \frac {max...}{(x-1)}\geq \frac {4(x+2)}{(x+2)}\Leftrightarrow \frac {max...}{(x-1)}\geq 4$

    My next step was this :

    $\displaystyle max\left \{ .... \right \}=x^{2}+x-2\Leftrightarrow x^{2}+x-2\geq 2x+4\Leftrightarrow x^{2}-x-6=(x+2)(x-3)\geq 0$

    So :
    $\displaystyle x+2\geq 0 \Rightarrow x\leq -2$
    $\displaystyle x-3\geq 0 \Rightarrow x\geq 3$

    So it follows:

    $\displaystyle max\left \{ x^{2}+x-2,2x+4 \right \}=\left\{\begin{matrix} x^{2}+x-2 &for & x\leq-2 & or & x\geq 3 \\ 2x+4&for &-2\leq x\leq 3 \end{matrix}\right.$

    and this is the point where I don't know how to continue. I mean, I have an example of it in my script but I don't know how to do that. Can someone help?
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  2. #2
    MHF Contributor

    Apr 2005
    Excellent! You have determined when $\displaystyle x^2+ x- 2$ is larger than $\displaystyle 2x+ 4$ and when $\displaystyle 2x+ 4$ is larger. Now just do this as three separate problems, on the three intervals, $\displaystyle x\le -2$, $\displaystyle -2\le x\le 3$, and $\displaystyle 3\le x$. In the first and last intervals, replace "$\displaystyle max\{x^2+ x- 2, 2x+4\}$" with $\displaystyle x^2+ x- 2$ and on the middle interval, replace it with $\displaystyle 2x+ 4$.
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  3. #3
    Newbie MysteriousPP's Avatar
    Dec 2010
    Ok I did what you said:

    If $\displaystyle \[x\leq -2\]$

    $\displaystyle \[\frac {x^{2}+x-2}{(x-1)}\geq 4\Leftrightarrow \frac {(x-1)(x+2)}{(x-1)}\geq 4 \Leftrightarrow x\leq 2\]$

    But what answer to give here ... I mean, how to form an answer here. The same goes with others

    If $\displaystyle \[x\geq 3\]$

    $\displaystyle \[\frac {x^{2}+x-2}{(x-1)}\geq 4\Rightarrow x\leq 2\]$ <-- but here I think it's totally false and impossible cause if x is higher than 3 then it can't be lower than 2 , right?

    If \[-2\leq x\leq 3\]

    $\displaystyle \[\frac {2x+4}{(x-1)}\geq 4\Leftrightarrow \frac {2(x+2)}{(x-1)}\geq 4 \mid * (x-1) \Leftrightarrow 2(x+2)\geq 4(x-1) \mid :2 \Leftrightarrow x+2\geq 2x-2\Leftrightarrow -x\geq -4\Leftrightarrow x\leq 4\]

    It all must be correct but I don't know how to form the answer hehehe :P I always hear that you can write whatever you want but I don't know...

    PS. And you basicly say that I need to take all of the intervals and do them as seperate problems? OK . Is it always so? Cause I have here an example, it looks very similar to this, but they also did a 4th problem using x=2 , and I have no idea why... :/
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