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Key Questions

  • The range of a function is its y-values or outputs. If you look at the graph from lowest point to highest point, that will be the range.

    Ex: #y = x^2# has a range of y#>=# 0 since the vertex is the lowest point, and it lies at (0,0).
    my screenshot

    Ex: y = 2x + 1 has a range from #-\infty# to #\infty# since the ends of the graph point in those directions. (down and left, and up and right)
    In interval notation, you would write #(-\infty,\infty)#.

    my screenshot2

    Ex: Some functions have interesting ranges like the sine function.
    y = sin(x) my screenshot3
    Its highest values are 1 and its lowest values are -1. That range is #-1<=y<=1# or [-1,1] in interval notation.

    Ex: A rather complicated function with a very challenging range is the inverse or reciprocal function, #y=frac{1}{x}#.

    my screenshot 4

    The output values might be difficult to describe except to say that they seem to include all real numbers except 0. (there is a horizontal asymptote on the x-axis)

    You could write #(-\infty,0)U(0,\infty)# in interval notation.

    Enjoy your study of range!

  • The range of a function is the set of all possible outputs of that function.

    For example, let's look at the function #y = 2x#

    Since we can plug in any x value and multiple it by 2, and since any number can be divided by 2, the output of the function, the #y# values, can be any real number.

    Therefore, the range of this function is "all real numbers"

    Let's look at something slightly more complicated, a quadratic in vertex form: #y=(x-3)^2+4#. This parabola has a vertex at #(3,4)# and opens upwards, therefore the vertex is the minimum value of the function. The function never goes below 4, therefore the range is #y>=4#.

  • Just like describing the domain of a function, you can use inequalities or interval notation; for example, you can write:

    Range: #[-2,3)# or #-2 le y < 3#

    I hope that this was helpful.


  • Jim H answered · 11 months ago
  • A08 answered · 1 year ago
  • Ken C. answered · 1 year ago