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Draw the graphs of some functions

Draw the graphs of the functions defined below on the interval [-2,2], and if it is a step function find a partition P such that the function is constant on the open subintervals of P.

  1. f(x) = x + [x].
  2. f(x) = x - [x].
  3. f(x) = [-x].
  4. f(x) = 2[x].
  5. f(x) = \left[x+\frac{1}{2} \right].
  6. f(x) = [x] + \left[ x+ \frac{1}{2} \right].

  1. This is not a step function. The graph is below.

    Rendered by QuickLaTeX.com

  2. This is not a step function. The graph is below.

    Rendered by QuickLaTeX.com

  3. This is a step function and it is constant on the open subintervals of the partition, P = \{ -2,-1,0,1,2 \}. The graph is below.

    Rendered by QuickLaTeX.com

  4. This is a step function and it is constant on the open subintervals of the partition, P = \{ -2,-1,0,1,2 \}. The graph is below.

    Rendered by QuickLaTeX.com

  5. This is a step function and it is constant on the open subintervals of the partition, P = \left\{ -2,-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2},2 \right\}. The graph is below.

    Rendered by QuickLaTeX.com

  6. This is a step function and it is constant on the open subintervals of the partition, P = \left\{ -2,-\frac{3}{2},-1,-\frac{1}{2},0,\frac{1}{2},1,\frac{3}{2},2 \right\}. The graph is below.

    Rendered by QuickLaTeX.com

7 comments

  1. Sebastian says:

    Dude i might be wrong but shouldn’t graph c be one to the right, cause de interval from -2 to -1 should be on 2 on the y axis.

  2. Yassin says:

    in th graph c should not be open on left side of each point of the partition and closed on the right side?

      • Yassin says:

        Thank you for this amazing web site, listen i am solving the book Introduction to Linear Algebra written by Serge Lang, if you want i can salve my solutions and send to you maybe can be usefull :) I am sorry if i wrote wrong some phrase, actually english is not my native language haha.

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