1. ## range of function

use any method to find the range of the function f, where
1. f(x) = 1/(x-2)(x-6)
2. f(x) = x/(1-x)^2

i really dont know where to start here. i could graph these functions, but i just don't manage to break them down to recognisable patterns. otherwise i thought i could find the domain of the inverse and then use that as the range, but that doesnt work does it? anyway i would be needing some help...

2. Originally Posted by furor celtica
use any method to find the range of the function f, where
1. f(x) = 1/(x-2)(x-6)
2. f(x) = x/(1-x)^2

i really dont know where to start here. i could graph these functions, but i just don't manage to break them down to recognisable patterns. otherwise i thought i could find the domain of the inverse and then use that as the range, but that doesnt work does it? anyway i would be needing some help...
to #1.:

$\displaystyle (f(x)<0\ \text{if}\ 2<x<6)~\vee~ (f(x)> 0\ \text{if}\ x < 2 \vee x>6)$

Now consider the case that x approaches 2:

$\displaystyle x\; \to 2 \wedge {x<2}~\implies~f(x) \to +\infty$ and $\displaystyle x\; \to 2 \wedge {x>2}~\implies~f(x) \to -\infty$

There exists a horizonzal asymptote y = 0 (Calculate $\displaystyle \lim_{|x| \to \infty} f(x)$ )

thus the range is $\displaystyle \mathbb{R} \setminus \{0\}$

to #2.

It could be done similarly. The change of the sign ocurrs this time in the numerator. I'll leave this part for you.

3. Originally Posted by furor celtica
use any method to find the range of the function f, where
1. f(x) = 1/(x-2)(x-6)
2. f(x) = x/(1-x)^2

i really dont know where to start here. i could graph these functions, but i just don't manage to break them down to recognisable patterns. otherwise i thought i could find the domain of the inverse and then use that as the range, but that doesnt work does it? anyway i would be needing some help...
Is this really a pre-calculus problem? - I wonder because the typical approach would be to determine the extreme values the function f assumes on any of the connected components (intervals) of its domain and then, by continuity of f on those components, conclude that f assumes all real values in between those extremes ("intermediate value theorem").

4. I have taught pre-calculus students how to find the range. It usually relies on a knowledge of the nature and shape of the functions in question.

5. Originally Posted by pickslides
I have taught pre-calculus students how to find the range. It usually relies on a knowledge of the nature and shape of the functions in question.
Well, I trust your experience (I have never taught pre-calculus: so I could not tell from my own experience). But this certainly works only for some functions. And it looks to me (I am sorry to say) that such problems, insofar as they are at all workable without knowledge of the derivative of a function, very likely are more like puzzles than something particularly important to do at this stage of the curriculum...

6. Originally Posted by Failure
But this certainly works only for some functions.
Agreed.

I think the range of the functions in question can be found without calculus.

7. ok i'm sure really heavy exchange was going on back there but i couldnt understand it so i cannot comment. however, the only post actually attempting to solve my problem was that by earboth which unfortunately:
1. utilized a method i am unacquainted with, which entails that a simpler solution must exist
2. was wrong, according to my textbook

so i am still awaiting help. thanks in advance.

8. Originally Posted by furor celtica
ok i'm sure really heavy exchange was going on back there but i couldnt understand it so i cannot comment. however, the only post actually attempting to solve my problem was that by earboth which unfortunately:
1. utilized a method i am unacquainted with, which entails that a simpler solution must exist
2. was wrong, according to my textbook

so i am still awaiting help. thanks in advance.
As to your first problem: the range does not change if you shift the entire graph horizontally. Since the Graph of f has vertical asymptotes at x=2 and x=6, I recomment shifting that graph to the left by 4 so that it comes to lie symmetrically with respect to the y-axis. So let $\displaystyle g(x) := f(x+4) =\frac{1}{(x+2)(x-2)}=\frac{1}{x^2-4}$.

Now consider first the case that $\displaystyle -2<x<2$: the largest value that g can assume for such an x is $\displaystyle g(0)=-4$. If you let x tend toward either -2 or +2 in this intervall (-2,2) you get arbitrarily large negative values of g. Since g is continuous it assumes all values inbetween, therefore the image of (-2,2) under g is $\displaystyle (-\infty,-4]$.
Next, consider the case that x<-2 or 2<x. Then, g(x)>0. You can get arbitrarily large values of g(x) for such an x close enough to -2 or +2. However, if you let x tend toward $\displaystyle +\infty$ or $\displaystyle -\infty$, the values of g stay positive but get ever closer to 0. Thus for x<-2 and 2<x the values of g can take any value in $\displaystyle (0,+\infty)$.
To sum up: the range of g - and thus the range of f - is $\displaystyle (-\infty,-4]\cup (0,+\infty)$.

9. Originally Posted by furor celtica
use any method to find the range of the function f, where
1. f(x) = 1/(x-2)(x-6)
2. f(x) = x/(1-x)^2

i really dont know where to start here. i could graph these functions, but i just don't manage to break them down to recognisable patterns. otherwise i thought i could find the domain of the inverse and then use that as the range, but that doesnt work does it? anyway i would be needing some help...
As to your second problem, consider that $\displaystyle f(x)=\frac{x}{(1-x)^2}=-\frac{1}{1-x}+\frac{1}{(1-x)^2}$. Set $\displaystyle u := \frac{1}{1-x}$ and note that u can assum any value in $\displaystyle \mathbb{R}\backslash\{0\}$. This gives $\displaystyle g(u) := -u+u^2$. What is the range of g and how is it related to the range of f?

10. but is there still not a simpler way of doing this, such as finding the inverse function? im sure i could get a better picture if i could graph this function, but it is too difficult for me. how can i do that using more basic notions?

11. Originally Posted by furor celtica
but is there still not a simpler way of doing this, such as finding the inverse function? im sure i could get a better picture if i could graph this function, but it is too difficult for me. how can i do that using more basic notions?
Sorry, but I have only so many ideas today .

You agree, I hope, that $\displaystyle g(u)=-u+u^2$ assumes the lowest value at $\displaystyle u=1/2$ and thus the range of g is $\displaystyle [-1/4,+\infty)$ (g is just a quadratic function, its graph is a parabola open from above). I maintain that the range of g is actually the same as the range of f. Thus, the range of f is $\displaystyle [-1/4,+\infty)$.