How do I determine the range of the function f(x) = (3x-4)/(3x+15)?

y = (3x - 4)/(3x + 15)

y(3x + 15) = 3x - 4

3xy + 15y = 3x - 4

3x(y - 1) = -4 - 15y

3x = (-4 - 15y)/(y - 1)

x = -(4 + 15y)/(3(y - 1))

So there is a real number x for any real number y except the case of y = 1.

Therefore, the range of y is (-inf,1)∪(1,inf).

y = (3x-4)/(3x+15)

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Range

x  =  -5 results in division by zero without making the numerator zero. So, there is a vertical

asymptote at x  =  -5. Therefore, y will be going to either +infinity or -infinity as x gets close

to -5.

      i. As x approaches -5 from the left, say   -6  <  x  <  -5, the numerator (3x-4) stays

                  negative and the denominator (3x+15) stays negative so that

                        (3x-4)/(3x+15)  ⇒  (neg)/(neg) = pos  ⇒  y goes to +inf.

                                                          y ➔ ∞

      ii. As x approaches -5 from the right, say   -5  <  x  <  -4, the the numerator (3x-4) stays

                  negative and the denominator (3x+15) stays positive so that

                        (3x-4)/(3x+15)  ⇒  (neg)/(pos) = neg  ⇒  y goes to -inf.

                                                          y ➔ −∞

      iii. As x gets larger and larger, i.e. as x goes to +inf, the -4 in the numerator and the

          +15 in the denominator become insignificant so that (3x-4)/(3x+15) approaches

            (3x)/(3x) or 1 ... But, note that it never gets to 1 so that 1 is not in the Range.

                                                              y  ≠  1

      iv. As x goes more and more negative, i.e. as x goes to -inf, the -4 in the numerator

          and the +15 in the denominator become insignificant so that (3x-4)/(3x+15) again

          approaches (3x)/(3x) or 1 ... But, note that, once again, it never gets to 1 so that

          1 is not in the Range.

                                                              y  ≠  1

Therefore, from i, ii, iii, and iv,

                  Range = { y  ∈  ℝ , y ≠ 1 }

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Domain

(3x-4)/(3x+15) is defined for all values of x except -5, because, -5 creates division by 0.

Therefore,

                  Domain = { x  ∈  ℝ , x ≠ -5 }

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The range of the function f(x) = (3x - 4)/(3x + 15):

{f element R : f!=1}

(assuming a function from reals to reals)