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

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

### 3 Answers

- atsuoLv 66 months 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].

- AmyLv 76 months 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.

- Wayne DeguManLv 76 months 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

:)>