# Thread: 3 Questions: 2 on Range and 1 on a logarithm

1. ## 3 Questions: 2 on Range and 1 on a logarithm

1. f(x)= (4-x^2)/(x^2-9)

What is the fuctions range?
_______

2. f(x)=(2x^2-3)/(x+1)

What is the fuctions range?
_______

3. When Log base b of A=2 and Log base b of D=5

What is: Log base b of (a+d)
_______

2. For #1,2, take $\displaystyle \lim_{x \to \pm \infty} f(x)$ and then see if there are any relative min/max.

3. ## Re:

We haven't learned how to take the "Limit" yet she wants us to look at the picture and tell. I know from looking at #3, that the function has a slant asymptote of y=2x-2. It would easy to find the range if the problem had a horizontal, but it doesn't. Like I said I haven't learned how to take the range yet, so there has to be some other way.

Thanks

4. Originally Posted by qbkr21
1. f(x)= (4-x^2)/(x^2-9)

What is the fuctions range?
Or you can use Hacker's trick.
By definition of range....
For what values of $\displaystyle y$ does the equation,
$\displaystyle y=\frac{4-x^2}{x^2-9}$ has a solution in the domain of the function?
Express as,
$\displaystyle -y=\frac{x^2-4}{x^2-9}=\frac{(x^2-9)+5}{x^2-9}=1+\frac{5}{x^2-9}$
Thus,
$\displaystyle -1-y=\frac{5}{x^2-9}$
If $\displaystyle y\not = -1$ then,
We have,
$\displaystyle \frac{1}{-1-y}=\frac{x^2-9}{5}$
Thus,
$\displaystyle \frac{5}{-1-y}=x^2-9$
Thus,
$\displaystyle \frac{9y+4}{1+y}=x^2$
To have a solution we require that,
$\displaystyle \frac{9y+4}{1+y}\geq 0$
Thus,
$\displaystyle 9y+4\geq 0 \mbox{ and }1+y\geq 0$
Or,
$\displaystyle 9y+4\leq 0 \mbox{ and }1+y\leq 0$
Solving the inequalities,
$\displaystyle y\geq -4/9 \mbox{ and }y\geq -1\mbox{ thus }y\geq -4/9$
Or,
$\displaystyle y\leq -4/9 \mbox{ and }y\leq -1\mbox{ thus }y<-1$ (remember $\displaystyle y\not = -1$)
Thus, the range is,
$\displaystyle (-\infty,-1)\cup [-4/9,+\infty)$
----
This is for Jameson, try to learn it my way. Because that method is graphical and works for "well-behaved" functions. If you want to do it mathematically this is the way.

5. Originally Posted by qbkr21
2. f(x)=(2x^2-3)/(x+1)

What is the fuctions range?
Again, for what values of $\displaystyle y$ does the function has a solution in the domain?
$\displaystyle y=\frac{2x^2-3}{x+1}$
Multiply by $\displaystyle (x+1)$ thus,
$\displaystyle xy+y=2x^2-3$
Thus,
$\displaystyle 2x^2-xy+(-3-y)=0$
This is a quadradic equation, it has real solutions when the discrimanant $\displaystyle b^2-4ac\geq 0$ thus,
$\displaystyle y^2-4(2)(-3-y)\geq 0$
Thus,
$\displaystyle y^2+8(3+y)\geq 0$
Thus,
$\displaystyle y^2+8y+24\geq 0$
Now we look at the discrimanat of this quadradic and we find that,
$\displaystyle 8^2-4(24)<0$ and $\displaystyle a>0$
Thus, this expression is always positive for any $\displaystyle y$. Hence the range is $\displaystyle (-\infty,+\infty)$

6. ## Re:

Sir I am completely lost how in the **** did you do that. What is "Hackers Trick"? This seems extreamly useful because you turned this problem into a enequality, which I can easily solve. What on earth did you do?

Thanks

7. Originally Posted by qbkr21
Sir I am completely lost how in the **** did you do that. What is "Hackers Trick"? This seems extreamly useful because you turned this problem into a enequality, which I can easily solve. What on earth did you do?

Thanks
Tell me exactly what you do not understand about the two problems?

How I got the inequalities?
You surly (CaptainBlank word) understood something.

8. ## Re

Sorry for using the 4 letter word you know as well as I do I simply forgot, my bad. Ok it seems to me you are finding the inverse or something of that nature. I don't really understand how you make Y negative and substitue it in for f(x), then I don't know what you are solving for, and then why are you setting it equal to zero? Maybe it a quadratic, yea that probably the reason. Some clarrification would be nice.

Thanks!

9. Originally Posted by qbkr21
Sorry for using the 4 letter word you know as well as I do I simply forgot, my bad. Ok it seems to me you are finding the inverse or something of that nature. I don't really understand how you make Y negative and substitue it in for f(x), then I don't know what you are solving for, and then why are you setting it equal to zero? Maybe it a quadratic, yea that probably the reason. Some clarrification would be nice.

Thanks!
Consider the function,
$\displaystyle f(x)=\frac{4-x^2}{x^2+9}$
Does what it mean the range?
It means all the possible values of the function.
For example, if $\displaystyle x=0$ then $\displaystyle f(0)=-4/9$, that means that $\displaystyle -4/9$ is in the range of $\displaystyle f(x)$.
So, I was doing this in general. Let $\displaystyle y$ be some real number. And I wanted to find what are the conditions that $\displaystyle y$ need to have so that I can find a value of $\displaystyle x$ such that,
$\displaystyle y=\frac{4-x^2}{x^2+9}$ meaning has a real solution (not imaginary).
Again, let me explain it. I am trying to find all the $\displaystyle y$ such that I can find an $\displaystyle x$ for it. That is what it means to be in the range of the function, because for some point $\displaystyle x$ we have $\displaystyle y=f(x)$ which we are tring to find.
After some manipulation (did you understand that?) we arrived at,
$\displaystyle \frac{9y+4}{y+1}=x^2$
Note, that this equation for $\displaystyle x$ cannot have a real solution when the left hand side is negative (square root of negative number, right?). Thus,
$\displaystyle \frac{9y+4}{y+1}\geq 0$.
We require that term to be positive.
When can that happen?
When both numberator and denominator are positive OR both numerator and denominator are negarive. (Two postitives divided are positive and two negatives divided are negative).
From there I solve the inequalities.

10. Originally Posted by ThePerfectHacker
Thus,
$\displaystyle \frac{9y+4}{1+y}=x^2$
To have a solution we require that,
$\displaystyle \frac{9y+4}{1+y}\geq 0$
Thus,
$\displaystyle 9y+4\geq 0 \mbox{ and }1+y\geq 0$
of course it wou;d really be
$\displaystyle 9y+4\geq 0 \mbox{ and }1+y> 0$

but it's possible for both parts to be negative and end with a positive outcome

EDIT: Just realized you took the negatives into account

11. Originally Posted by Quick
of course it wou;d really be
$\displaystyle 9y+4\geq 0 \mbox{ and }1+y> 0$
I understand what you are trying to say since ($\displaystyle y\not = -1$). But that is not such a big problem for I already mentioned that. You can do that if you really want.

but it's possible for both parts to be negative and end with a positive outcome
Yes,
$\displaystyle \frac{-2}{-2}=1\geq 0$

12. Originally Posted by ThePerfectHacker
I understand what you are trying to say since ($\displaystyle y\not = -1$). But that is not such a big problem for I already mentioned that. You can do that if you really want.
Well, I was going for the fact that the denominator couldn't be zero, but I guess it works out in the end...

Yes,
$\displaystyle \frac{-2}{-2}=1\geq 0$
Yes I am quite aware, I used to show that I was wondering why you left that out of your solution, but it turns out that I was skiffing through and missed where you said "or"

13. ## Re:

First of all how are you guys able to type that mathmatical text into the computer. I am unable to do so. Secondly I put Y is not equal to -1 on the test and she marked it wrong. I only missed these 3 problems on a 50 question test. I got an A+ however, as you are well aware I do have some questions.

Thanks

14. ## RE: Manipulation

How did you manipulate the original problem to get:

(9y+4)/(y+1)=x^2

You obviously did multiply by the denominator of the original problem because that would give you this answer.

Thanks

15. Originally Posted by qbkr21
First of all how are you guys able to type that mathmatical text into the computer. I am unable to do so. Secondly I put Y is not equal to -1 on the test and she marked it wrong. I only missed these 3 problems on a 50 question test. I got an A+ however, as you are well aware I do have some questions.

Thanks

Then type stuff in between the tags like: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

to get: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

there is a tutorial in the "announcements" forum (the one on the bottom of the main page)

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