Results 1 to 10 of 10

Math Help - Max and min of absolute value interval

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    31

    Max and min of absolute value interval

    Let f(t)=3-|t-3| on the interval [-1,5]. Find the absolute maximum and absolute minimum of f(t) on this interval.

    The absolute max occurs at t = 3
    The absolute min occurs at  t = -1

    I think I am doing this wrong. Those are the correct answers but I have no idea how 3 is the max. fuggled to me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by pseizure2000 View Post
    Let f(t)=3-|t-3| on the interval [-1,5]. Find the absolute maximum and absolute minimum of f(t) on this interval.

    The absolute max occurs at t = 3
    The absolute min occurs at  t = -1

    I think I am doing this wrong. Those are the correct answers but I have no idea how 3 is the max. fuggled to me.
    why does that "fuggle" you? since whatever is in absolute value signs is nonneggative, the maximum value of the function is when what is in the absolute value signs is zero. so it would be 3 - 0 = 3. this happens on the interval when t = 3

    what specifically is your concern? how did you work out the problem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    I'd do it like this---I will test the f(t) for every t in the given interval.

    f(t) = 3 -|t -3|, in the interval [-1,5]

    f(-1) = 3 -|-1 -3| = -1 <--------------minimum
    f(0) = 3 -|0 -3| = 0
    f(1) = 3 -|1 -3| = 1
    f(2) = 3 -|2 -3| = 2
    f(3) = 3 -|3 -3| = 3 <-----------------maximum
    f(4) = 3 -|4 -3| = 2
    f(5) = 3 -|5 -3| = 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ticbol View Post
    I'd do it like this---I will test the f(t) for every t in the given interval.

    f(t) = 3 -|t -3|, in the interval [-1,5]

    f(-1) = 3 -|-1 -3| = -1 <--------------minimum
    f(0) = 3 -|0 -3| = 0
    f(1) = 3 -|1 -3| = 1
    f(2) = 3 -|2 -3| = 2
    f(3) = 3 -|3 -3| = 3 <-----------------maximum
    f(4) = 3 -|4 -3| = 2
    f(5) = 3 -|5 -3| = 1
    well, t does not have to be an integer it seems. t can be a real number, in which case someone could argue that: "perhaps a t you neglected to check gives the absolute max"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Jhevon View Post
    well, t does not have to be an integer it seems. t can be a real number, in which case someone could argue that: "perhaps a t you neglected to check gives the absolute max"
    I see.

    Okay, say t = 2.99
    f(2.99) = 3 -|2.99 -3|= 2.99

    Say again t = 3.001
    f(3.001) = 3 -|3.001 -3| = 2.999

    So, where would f(t) be greater than 3 in the given interval?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2007
    Posts
    31
    fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

    maybe i'm not doing it right but I couldn't find 3 as a critical point
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by pseizure2000 View Post
    fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

    maybe i'm not doing it right but I couldn't find 3 as a critical point
    your instructor is correct, the thing is, f'(x) does not exist at 3 (the graph is a cusp at t = 3, we have the same problem when we try to evaluate the derivative of |x| at x = 0, for instance), that's why you can't find it. you are dealing with absolute values here, you need to split the interval in two and work on it piece by piece. find where the turning point is, and then split the interval at that turning point. here the turning point is 3, so consider the interval [-1,3] and [3,5]. on both interval, the absolute max is 3, so we conclude that the absolute max is 3 for the entire interval
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by pseizure2000 View Post
    fuggled because my teach told me i need to evaluate all critical points (where f'(x) DNE and/or f'(x) = 0 and endpoints

    maybe i'm not doing it right but I couldn't find 3 as a critical point
    I suspected using the first derivative of f(t) is one way, but I forgot how to differentiate an absolute quantity.

    So I thought of doing it graphically.
    The graph of f(t) = |t| is a V that opens upward, whose vertex is at (0,0), and the slopes of the two rays are -1 or 1, or they are at 45 degrees with the horizontal.

    You make that f(t) = -|t|, then you "invert", or turn upside down, the graph of f(t) = |t|

    Then you add 3 to the -|t| to make it f(t) = 3 -|t|, and you shift the vertex of f(t) = -|t| vertically upward by 3 units.

    Then you subtract 3 from the |t| to make it |t -3|, and you shift the graph 3 units to the right.

    You graph the final f(t) = 3 -|t-3|. You'd see that its vertex is a maximum and it is at (3,3).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    We have -1\leq t\leq 5.
    Substract 3 from all members: -4\leq t-3\leq 2\Rightarrow 0\leq|t-3|\leq 4.
    Multyply all members by -1: 0\geq-|t-3|\geq -4.
    Add 3 to all members: 3\geq 3-|t-3|\geq -1.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by red_dog View Post
    We have -1\leq t\leq 5.
    Substract 3 from all members: -4\leq t-3\leq 2\Rightarrow 0\leq|t-3|\leq 4.
    Multyply all members by -1: 0\geq-|t-3|\geq -4.
    Add 3 to all members: 3\geq 3-|t-3|\geq -1.
    Very nice, simple yet very impressive, as always
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Express an interval with an absolute value
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 11th 2012, 09:47 AM
  2. Replies: 8
    Last Post: May 23rd 2010, 11:59 PM
  3. Absolute Max and Min on an Interval? =[
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 5th 2010, 09:12 PM
  4. Replies: 4
    Last Post: September 3rd 2009, 05:22 PM
  5. Replies: 2
    Last Post: February 11th 2007, 08:16 AM

Search Tags


/mathhelpforum @mathhelpforum