1. ## domain and range

Can someone explain domain and range to me? I recall doing these problems this past year, but since all textbooks were turned in in May, I have no point of reference for my summer homework, and my memory is a little foggy.

The first problem says find the domain of the function: f(x)= (square root of (x-5))/(x^2-6x+8)

The other problem says find the domain and range of the funtion:
f(x)=(4/5)l8-xl

A refresher course in domain and range would be helpful.

2. Originally Posted by iheartthemusic29
The first problem says find the domain of the function: f(x)= (square root of (x-5))/(x^2-6x+8)
$\displaystyle f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}$

What are exactly the conditions you must set so the function is well defined?

3. Originally Posted by iheartthemusic29
The other problem says find the domain and range of the funtion:
f(x)=(4/5)l8-xl
$\displaystyle f(x)=\frac45\left|8-x\right|$

How'd you find the domain?

Having an absolute value function, what could you conclude about the range?

4. Originally Posted by Krizalid
$\displaystyle f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}$

What are exactly the conditions you must set so the function is well defined?
What do you mean? I gave all the information I have...after that, I know nothing.

5. Sorry, I didn't read the first

Well, in few words... domain means to find the values of $\displaystyle x$ for which the function is well defined.

For example, in your first problem, you can't have a negative root, so one condition would be setting $\displaystyle x-5\ge0$, you can set it equal to zero 'cause numerator can be zero.

But the denominator can't be zero, so you must set $\displaystyle x^2-6x+8\ne0$ (got the idea?)

About the range, are all possible $\displaystyle y$ values that the function can take.

Sorry for this useless explanation but I have more things to do

6. Originally Posted by iheartthemusic29
The first problem says find the domain of the function: $\displaystyle f(x)= \sqrt{\frac{x-5}{x^2-6x+8}}$
Domains are fairly easy. Typical things you want to do is check for any of the following:
1) Places where you get a 0 under a fraction
2) Where you would get a negative under a square root
These are the big ones, and you have both in this problem.

Ranges are tougher. ThePerfectHacker has a method for doing these (go figure that I couldn't get the search engine to find any of his examples). Probably the simplest way is simply to graph it and see what happens.

In your case we get a zero in the denominator for
$\displaystyle x^2 - 6x + 8 = 0$

$\displaystyle (x - 2)(x - 4) = 0$

So the points x = 2 and x = 4 are out of the domain.

We also want to see where the fraction
$\displaystyle \frac{x-5}{x^2-6x+8} = \frac{x - 5}{(x - 2)(x - 4)}$
is negative. We search for where the numerator and denominator are 0. This is at x = 5, x = 2, and x = 4. (These are called "critical points.") Now split up the real number line and ask the question:
$\displaystyle ( -\infty, 2 )$: $\displaystyle \frac{x - 5}{(x - 2)(x - 4)} < 0$

$\displaystyle ( 2, 4 )$: $\displaystyle \frac{x - 5}{(x - 2)(x - 4)} > 0$

$\displaystyle ( 4, 5 )$: $\displaystyle \frac{x - 5}{(x - 2)(x - 4)} < 0$

$\displaystyle ( 5, \infty )$: $\displaystyle \frac{x - 5}{(x - 2)(x - 4)} > 0$

So we have the conditions on the domain: x cannot be 2 or 4, and cannot be in the intervals $\displaystyle (-\infty, 2)$ nor (4, 5). The domain is everything else.

So the domain of the function is $\displaystyle (2, 4) \cup [5, \infty)$. (We include x = 5 because there is no reason we can't have a 0 under the square root.)

For the range, look at the graph below. It looks to me like the range will be $\displaystyle [0, \infty)$.

-Dan