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Math Help - Greatest integer function (=step function) problem

  1. #1
    s3a
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    Greatest integer function (=step function) problem

    In this question it tells me to find the zeros which means that f(x) = 0. However my problem is that, I do not know how to undo the [] brackets. Lets say we were given the following square root function and we wanted to find its zero:

    f(x) = sqrt(x)
    0 = sqrt(x)
    0^2 = (sqrt(x))^2
    0 = x
    x = 0
    (notice how the square root was undone by squaring sqrt(x))

    for the step function:

    f(x) = 3[-(x-1)/2] + 6
    0 = 3[-(x-1)/2] + 6
    -6/3 = [-(x-1)/2]
    -2 = [-(x-1)/2]

    ^^ Here is where I am stuck since I do not know how to undo the [] brackets.^^

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Clearly you need \left[ { - \frac{{x - 1}}{2}} \right] =  - 2.
    Do you understand that \left[ { - \frac{3}{2}} \right] =  - 2?
    Also \left[ { -{2}} \right] =  - 2.
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  3. #3
    s3a
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    I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

    [-(x-1)/2] = -2
    [-(x-1)] = -4
    [x-1] = 4
    [x] = 5

    That's the furthest I can go. is that even right? The answer seems so off compared to the ]3, 5] that I am supposed to get!

    Please help!
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  4. #4
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    Quote Originally Posted by s3a View Post
    I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

    [-(x-1)/2] = -2
    [-(x-1)] = -4

    [x-1] = 4
    [x] = 5
    That is easy to do.
    The floor function, greatest integer function, does work that way.
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  5. #5
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    Quote Originally Posted by s3a View Post
    The answer seems so off compared to the ]3, 5] that I am supposed to get!
    Here is how it works.
     \begin{gathered}<br />
   - 2 \leqslant  - \frac{{x - 1}}<br />
{2} <  - 1 \hfill \\<br />
   - 4 \le  - x + 1 <  - 2 \hfill \\<br />
  3 < x \leqslant 5 \hfill \\ <br />
\end{gathered} .
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  6. #6
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    Quote Originally Posted by s3a View Post
    I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

    [-(x-1)/2] = -2
    [-(x-1)] = -4
    [x-1] = 4
    [x] = 5

    That's the furthest I can go. is that even right? The answer seems so off compared to the ]3, 5] that I am supposed to get!
    [x]= n, for n an integer, is true for any x between n and n+1.
    It is NOT true that if [-(x-1)/2]= -2 then [-(x-1)]= -4. For example if x= 3.5, then -(x-1)= -(2.5)= -2.5. so -(x-1)/2= -1.25 so [-(x-1)/2]= -2. But [-(x-)]= -3, not -4.

    Better to think: if [y]= -2 then -2\le y< -1. Now replace y by -(x-1)/2: -2\le -(x-1)/2< -1 so -4\le -x+1< -2, -5\le -x< -3, 3< x\le 5

    Blast you, Plato! Again, 2 minutes ahead of me!

    Please help!
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