# Thread: Finding domain and range.

1. ## Finding domain and range.

for the equation y=root 4-x^2

how do you find the range and domain???

for the equation x^2-9x+20 / x-5

how do you find the range???

2. Generally, the most effective method for finding the domain is to assume "All Real Numbers" and then see if you can rule out anything.

In the first, 4 - x^2 needs to be positive. Can you determine where it is negative?
In the first, x^2 is always positive. What's the greatest value that 4 - x^2 can take?

You must think about these. There is not a magic formula.

3. Originally Posted by maigowai

for the equation x^2-9x+20 / x-5
Here's a hint $\frac{x^2-9x+20}{ x-5} = \frac{(x-5)(x-4)}{ x-5}$

4. Originally Posted by maigowai
for the equation y=root 4-x^2

how do you find the range and domain???

for the equation x^2-9x+20 / x-5

how do you find the range???
I find the best way of finding the range is to first try and draw a graph of the function and then get it from the graph. But as has already been said, there is no magic formula, each case has to be considered on its own merit.

5. Originally Posted by pickslides
Here's a hint $\frac{x^2-9x+20}{ x-5} = \frac{(x-5)(x-4)}{ x-5}$
That strikers me as NOT being a good hint for finding the domain! One might, mistakenly, cancel the "x- 5" factors and think the domain is "all real numbers"- which it is NOT.
$f(x)= \frac{(x-5)(x-4)}{x-5}$ is NOT the same as f(x)= x- 4.

6. Originally Posted by HallsofIvy
That strikers me as NOT being a good hint for finding the domain! One might, mistakenly, cancel the "x- 5" factors and think the domain is "all real numbers"- which it is NOT.
$f(x)= \frac{(x-5)(x-4)}{x-5}$ is NOT the same as f(x)= x- 4.
Minor nitpick, but the OP did not ask for the domain, he/she asked for the range. Nevertheless, your point still holds, the notion of canceling the "x-5" factors. However, knowing that the graph of the original f(x) is almost the same as g(x) = x - 4 (I'm calling this g(x)) is helpful to find the range. The graph of both f(x) and g(x) are lines, except for at x = 5. At x = 5, f(x) is not defined but g(x) = 1. That tells me that the range of f(x) is $( -\infty, 1) \cup (1, \infty)$.