Results 1 to 10 of 10

Thread: Interval Form?

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    41

    Interval Form?

    how do i write this in interval form??

    (x+2)/(x-5)≤0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    You should first note that $\displaystyle x \neq 5$.



    $\displaystyle \frac{x + 2}{x - 5} \leq 0$

    $\displaystyle 1 + \frac{7}{x - 5} \leq 0$

    $\displaystyle \frac{7}{x - 5} \leq 1$


    Now we will have two alternatives, depending on whether the denominator is positive or negative...

    Case 1: $\displaystyle x - 5$ is positive, which means $\displaystyle x > 5$.


    $\displaystyle 7 \leq x - 5$

    $\displaystyle 14 \leq x$

    $\displaystyle x \geq 14$.



    Case 2: $\displaystyle x - 5$ is negative, which means $\displaystyle x < 5$.

    $\displaystyle 7 \geq x - 5$

    $\displaystyle 14 \geq x$

    $\displaystyle x \leq 14$.


    It's pretty obvious that the ONLY $\displaystyle x$ which can not satisfy this equation are $\displaystyle x = 5$.


    So the domain is $\displaystyle x < 5 \cup x > 5$.

    Alternatively, it can be written $\displaystyle x \neq 5$ or $\displaystyle (-\infty , 5) \cup (5, \infty)$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    5
    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    please be specific. are we talking about writing the domain in interval notation or the solution to the inequality in interval notation?

    I suspect the latter. In which case set the numerator and denominator to zero. You get x = -2 or x = 5 (now note that x cannot be 5, because the denominator cannot in fact be zero, we just do this to figure out the signs). Now you can draw a number line and plot x = -2 and x = 5 on it. Then plug in numbers in each of the three intervals that you have. you will find that the middle interval is the only one that works. the resulting solution is $\displaystyle [-2, 5)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,801
    Thanks
    3035
    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    A fraction is negative if and only if numerator and denominator are of different signs. That is:

    1) $\displaystyle x+2\ge 0$ and x- 5< 0
    which lead to $\displaystyle x\ge -2$ and x< 5: [-2, 5)
    or

    2) $\displaystyle x+2\le 0$ and x- 5> 0
    which lead to $\displaystyle x\le -2$ and x> 5 which are impossible together.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    41
    Quote Originally Posted by Prove It View Post
    You should first note that $\displaystyle x \neq 5$.



    $\displaystyle \frac{x + 2}{x - 5} \leq 0$

    $\displaystyle 1 + \frac{7}{x - 5} \leq 0$

    $\displaystyle \frac{7}{x - 5} \leq 1$


    Now we will have two alternatives, depending on whether the denominator is positive or negative...

    Case 1: $\displaystyle x - 5$ is positive, which means $\displaystyle x > 5$.


    $\displaystyle 7 \leq x - 5$

    $\displaystyle 14 \leq x$

    $\displaystyle x \geq 14$.



    Case 2: $\displaystyle x - 5$ is negative, which means $\displaystyle x < 5$.

    $\displaystyle 7 \geq x - 5$

    $\displaystyle 14 \geq x$

    $\displaystyle x \leq 14$.


    It's pretty obvious that the ONLY $\displaystyle x$ which can not satisfy this equation are $\displaystyle x = 5$.


    So the domain is $\displaystyle x < 5 \cup x > 5$.

    Alternatively, it can be written $\displaystyle x \neq 5$ or $\displaystyle (-\infty , 5) \cup (5, \infty)$.
    ok do i put any brackets in the interval form????? thats mainly what im confused about. when do i use brackets and when do i not?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2009
    Posts
    41
    Quote Originally Posted by HallsofIvy View Post
    A fraction is negative if and only if numerator and denominator are of different signs. That is:

    1) $\displaystyle x+2\ge 0$ and x- 5< 0
    which lead to $\displaystyle x\ge -2$ and x< 5: [-2, 5)
    or

    2) $\displaystyle x+2\le 0$ and x- 5> 0
    which lead to $\displaystyle x\le -2$ and x> 5 which are impossible together.
    no disrespect but i think your wrong. my book says the answer is (-infin, 5] U [5, infin)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    no, he's correct, and i do think you should reconsider twice on determining when an answer is wrong or not, see the procedure first.

    i wouldn't be surprised of another typo on a book, so in this case, book's answer is wrong and HallsofIvy is correct.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    HallsofIvy's answer is indeed correct. To demonstrate why $\displaystyle (-\infty,5)\cup(5,\infty)$ is wrong, try substituting $\displaystyle x=6$ into the inequality and see what you get. Actually, any $\displaystyle x>5$ gives a positive result, in contradiction to this answer.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    hello
    if we take the first case of what "Hallsofivy" posted.
    we'll have $\displaystyle -2\leq x< 5$ as a solution.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,801
    Thanks
    3035
    Quote Originally Posted by Raoh View Post
    hello
    if we take the first case of what "Hallsofivy" posted.
    we'll have $\displaystyle -2\leq x< 5$ as a solution.
    However, IF the original question was "what is the domain of the function $\displaystyle \frac{x+2}{x- 5}$, THEN the answer is $\displaystyle (-\infty, 5)\cup(5, \infty)$. Prove It said that in his first response!

    However, I don't see what the "<=" has to do with it then. You find domains of functions, not inequalities.
    Last edited by mr fantastic; Nov 16th 2009 at 03:49 AM. Reason: Fixed some latex tags
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 11th 2010, 11:39 AM
  2. Interval form??
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Nov 15th 2009, 05:57 PM
  3. Replies: 10
    Last Post: Jul 14th 2009, 03:26 PM
  4. Replies: 14
    Last Post: May 30th 2008, 06:10 AM
  5. Increasing interval and decreasing interval
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 24th 2006, 08:49 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum