Results 1 to 5 of 5

Math Help - Basic "solve for x" problem

  1. #1
    Newbie
    Joined
    Jun 2011
    Posts
    12

    Basic "solve for x" problem

    "If f(x) = x^2 - 3x, determine the values of x for which f(x - 3) \leq 0"

    This is for only two marks, but I'm not quite sure how to go about doing this... Any tips would be great!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    May 2010
    Posts
    1,028
    Thanks
    28

    Re: Basic "solve for x" problem

    f(x-3) = (x-3)^2 - 3(x-3) = (x-3)^2 - 3x +9

    Can you find the values of X for which the above expression is \leq 0

    hint: Write it as a quadratic then find the solutions that give exactly 0. Then plot the graph and the range of values is clear

    Spoiler:

    http://www.wolframalpha.com/input/?i=%28x-3%29^2+-+3x+%2B9+%3C%3D+0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2011
    Posts
    12

    Re: Basic "solve for x" problem

    Got it, thank you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428

    Re: Basic "solve for x" problem

    Quote Originally Posted by SpringFan25 View Post
    f(x-3) = (x-3)^2 - 3(x-3) = (x-3)^2 - 3x +9

    Can you find the values of X for which the above expression is \leq 0

    hint: Write it as a quadratic then find the solutions that give exactly 0. Then plot the graph and the range of values is clear

    Spoiler:

    http://www.wolframalpha.com/input/?i=%28x-3%29^2+-+3x+%2B9+%3C%3D+0
    Or if you didn't want to waste time graphing, completing the square solves that quadratic inequality beautifully...

    \displaystyle \begin{align*}(x - 3)^2 - 3x + 9 &\leq 0 \\ x^2 - 6x + 9 - 3x + 9 &\leq 0 \\ x^2 - 9x + 18 &\leq 0 \\ x^2 - 9x + \left(-\frac{9}{2}\right)^2 - \left(-\frac{9}{2}\right)^2 + 18 &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{72}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{9}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 &\leq \frac{9}{4} \\ \left|x - \frac{9}{2}\right| &\leq \frac{3}{2} \\ -\frac{3}{2} \leq x - \frac{9}{2} &\leq \frac{3}{2} \\ 3 \leq x &\leq 6  \end{align*}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2011
    Posts
    4

    Re: Basic "solve for x" problem

    whoops... disregard. I misinterpreted the original question... My bad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 6th 2011, 03:00 PM
  2. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  3. Replies: 1
    Last Post: June 4th 2010, 10:26 PM
  4. Basic "Solve for X" with arctan
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: November 9th 2009, 10:11 PM
  5. Why is my "basic ratio" & "Z-score difference" different?!
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 7th 2008, 04:29 AM

Search Tags


/mathhelpforum @mathhelpforum