domain and range

**iheartthemusic29** · July 17th 2007, 12:16 PM

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.

**Krizalid** · July 17th 2007, 12:57 PM

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)

$f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}$

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

**Krizalid** · July 17th 2007, 01:01 PM

Originally Posted by iheartthemusic29

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

$f(x)=\frac45\left|8-x\right|$

How'd you find the domain?

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

**iheartthemusic29** · July 17th 2007, 01:02 PM

Originally Posted by Krizalid

$f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}$

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.

**Krizalid** · July 17th 2007, 01:09 PM

Sorry, I didn't read the first

Well, in few words... domain means to find the values of $x$ 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 $x-5\ge0$ , you can set it equal to zero 'cause numerator can be zero.

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

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

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

**topsquark** · July 17th 2007, 01:34 PM

Originally Posted by iheartthemusic29

The first problem says find the domain of the function: $f(x)= \sqrt{\frac{x-5}{x^2-6x+8}}$

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
$x^2 - 6x + 8 = 0$

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

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

We also want to see where the fraction
$\frac{x-5}{x^2-6x+8} = \frac{x - 5}{(x - 2)(x - 4)}$
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:
$( -\infty, 2 )$ : $\frac{x - 5}{(x - 2)(x - 4)} < 0$

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

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

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

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

So the domain of the function is $(2, 4) \cup [5, \infty)$ . (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 $[0, \infty)$ .

-Dan

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