How do I find the range of the function y=ax+d/cx-d without calculating what the domain of the inverse is?

Update:

Thanks for the quick reply, Wayne DeguMan. However, I’m still not getting your math nor explanation. Can you elaborate please?

3 Answers

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  • atsuo
    Lv 6
    1 month ago

    "y=ax+d/cx-d" means y = ax + (d/c)x - d, so I think you should use parentheses as y = (ax + d)/(cx - d).

    Case1. c ≠ 0

    y = (ax + d)/(cx - d)

     = [(a/c)(cx - d) + ad/c + d]/(cx - d)

     = a/c + (ad/c + d)/(cx - d)

     = a/c + d(a/c + 1)/(cx - d)

    The denominator (cx - d) can become any value.

    If a/c ≠ -1 and d ≠ 0 then d(a/c + 1) ≠ 0, so d(a/c + 1)/(cx - d) can become any value except 0. Therefore, y can become any value except a/c.

    The range is (-inf,a/c)∪(a/c,+inf).

    If a/c = -1 or d = 0 then d(a/c + 1) = 0, so y can become only a/c.

    The range is [a/c,a/c].

    Case2. c = 0 (The denominator (cx - d) must not be 0, so d ≠ 0)

    y = (ax + d)/(cx - d)

     = (ax + d)/(-d)

     = -(a/d)x - 1

    If a ≠ 0 then y can become any value.

    The range is (-inf,+inf).

    If a = 0 then y can become only -1.

    The range is [-1,-1].

  • Amy
    Lv 7
    1 month ago

    Suppose we want to know whether some value k is in the range.

    This is the same as asking whether there is a value for x such that k = (ax+d)/(cx-d).

    Solve for x:

    x = (k+1)d/(kc-a)

    Any value of k can produce some value for x, except for k = a/c.

    Thus the range of y is all values except a/c.

  • 1 month ago

    When considering the range, we are looking for values that the function cannot take, i.e. values of y that don't exist

    Dividing by x we get:

    y = (a + (d/x))/(c - (d/x))

    As x --> ± ∞, d/x --> 0

    so, y --> a/c

    Hence, range is all values except y = a/c

    In other words, as we increase the values of x, the value of the function y

    gets closer and closer to a value of a/c.

    e.g. if we had y = (2x + 1)/(3x - 1), we could then say:

    y = (2 + (1/x))/(3 - (1/x))

    Again, as x gets big, 1/x = 0

    so, y approaches 2/3...but never actually gets there.

    A sketch is below, where the dotted line is the value y = 2/3

    :)>

    Attachment image
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