1. ## range of function?

I'm having trouble figuring the range of this function:

y = [square root] (x-2) / (6+x) [/square root]

I know that the horizontal asymptote is y = 1, so there would be some restrictions there. I am just not sure how to use that info to find the range.

Thank you!

The answers are y > 0 and y < 1 .. I graphed it and I can see how it works out, just need some explanation please

2. Originally Posted by finalfantasy
I'm having trouble figuring the range of this function:

m(u) = [square root] (u-2) / (6+u) [/square root]

I know that the horizontal asymptote is y = 1, so there would be some restrictions there. I am just not sure how to use that info to find the range.

Thank you!
do you mean $m(u) = \sqrt{\frac {u - 2}{6 + u}}$ ?

remember, the range is all the possible outputs we have. what are the possible outputs if $u \in (- \infty, 6-) \cup [2, \infty)$? (assuming of course, we are dealing with real valued functions here.)

EDIT: if the above is your function, then we do not require that y < 1. and by the way, there is no y here. there is u and there is m

3. Yes, I mean that equation

4. Originally Posted by finalfantasy
Yes, I mean that equation
well, the range is $[0, \infty)$ graph it again, or try and tell me why through explanation

maybe you did somethign wrong when graphing by hand. graph it with this

5. Originally Posted by Jhevon
well, the range is $[0, \infty)$ graph it again, or try and tell me why through explanation

maybe you did somethign wrong when graphing by hand. graph it with this
Wouldn't the range be: $[0,1]$ because the function $\frac{x-2}{x+6}$ has an infinite limit of one giving it a horizontal asymptote of $y=1$.

6. Originally Posted by colby2152
Wouldn't the range be: $[0,1]$ because the function $\frac{x-2}{x+6}$ has an infinite limit of one giving it a horizontal asymptote of $y=1$.
horizontal asymptotes tell how the graph behaves at the "ends", that is, as $x \to \infty$ and $x \to - \infty$. what about in the middle? how big can the function get as $x \to -6$, for instance?

see the graph below. indeed, the horizontal asymptote is 1, but as $x \to -6$ (from the left, of course), $\sqrt{ \frac {x - 2}{x + 6}} \to \infty$

7. True, I had a mental block there from looking at the negative values - it's that whole square root thing!

8. Originally Posted by colby2152
Wouldn't the range be: $[0,1]$ because the function $\frac{x-2}{x+6}$ has an infinite limit of one giving it a horizontal asymptote of $y=1$.
Let $g(x) = \frac{x - 2}{x + 6} = \frac{x + 6 - 8}{x + 6} = 1 - \frac{8}{x + 6}$. g has a range of (-oo, 1) U (1, +oo).

Let $f(x) = \sqrt{x}$. The maximal domain of f is [0, +oo).

Therefore $y = \sqrt{\frac{x-2}{x+6}} = f(g(x))$ is only defined when the range of g is resticted to lie in [0, oo).
Under this restriction, the 'maximal input' for f will be [0, 1) U (1, +oo) and so the 'maximal output' of f will be [0, 1) U (1, +oo).

Therefore the range of $y = \sqrt{x-2}{x+6}$ is [0, 1) U (1, +oo).

9. Originally Posted by mr fantastic
Let $g(x) = \frac{x - 2}{x + 6} = \frac{x + 6 - 8}{x + 6} = 1 - \frac{8}{x + 6}$. g has a range of (-oo, 1) U (1, +oo).

Let $f(x) = \sqrt{x}$. The maximal domain of f is [0, +oo).

Therefore $y = \sqrt{\frac{x-2}{x+6}} = f(g(x))$ is only defined when the range of g is resticted to lie in [0, oo).
Under this restriction, the 'maximal input' for f will be [0, 1) U (1, +oo) and so the 'maximal output' of f will be [0, 1) U (1, +oo).

Therefore the range of $y = \sqrt{x-2}{x+6}$ is [0, 1) U (1, +oo).
yes, 1 is not in the range. thanks for pointing out that oversight