# Help me find the range of an equation please

1. Jul 24, 2010

### GreenPrint

1. The problem statement, all variables and given/known data
FIXED :)

y = (x^2 -4)/(x^2 - x -12)

dy/dx of numerator I got 2x
dy/dx of denomenator I got 2x - 1
dy/dx of y I got -(x^2 - 16x -4)/((x-4)^2(x+3)^2)

setting numerator to zero and solving got 8 +/- sqrt(65) with the critical value being the +
setting denomenator to zero and solving got 4 and -3 which are both out of range of the original function...

so now what how do I find the range kind of confused... found one critical value how do I find the others kind of lost here thanks

2. Relevant equations

3. The attempt at a solution

Last edited: Jul 24, 2010
2. Jul 24, 2010

### jegues

Don't you already have a thread with the exact same question in the precalculus section?

3. Jul 24, 2010

### GreenPrint

Yes then I realized after reading comments that it shouldn't be in there becaue it's now calculus stuff... can you help me?

4. Jul 24, 2010

### Staff: Mentor

That's not a good reason to post the same topic in another section.

5. Jul 24, 2010

### GreenPrint

Fixed error in thsi post as well ok :)

I'm tyring to find the range of the following equation

(x^2 - 4)/(x^2 - x -12)

In order to do so I have found the derivitive

-(x^2 + 16x + 4)/(x^2-x-12)^2

setting the numerator to zero and solving yields to extrema 2(4 +/- sqrt(15)) the only probelm I have is that a quick check of the graph shows me that there are three how do I find the other one that I can't get... it's the one in the fourth quardrant whose line appears only in the first quadrant, yes I know it's all one line, on the graph, i.e. not the one in the first or the one in all four...

pleaes show me how to do this or tell me thanks

Last edited: Jul 24, 2010
6. Jul 24, 2010

### vela

Staff Emeritus
You didn't differentiate the function correctly. Is the denominator correct? It seems a bit strange to have two linear terms, and I don't see how you got the denominator of the derivative from the denominator of the original function.

EDIT: Never mind. I see you have a different function in the original post of this thread. The derivative is close but doesn't quite match what Mathematica gives, so check your algebra. The quadratic term in the numerator should be negative.

7. Jul 24, 2010

### GreenPrint

Yes I believe it's correct the quotient rule says that the denomenator of the derivitive is the denomenator squared of the original funciton... no? I thought so I don't know I must be doing something wrong can't figure it out

8. Jul 24, 2010

### GreenPrint

http://en.wikipedia.org/wiki/Quotient_rule
see but ya I must be doing something wrong don't see what the numeraotr I believe gave me two correct ones just don't know how to find the third critical value don't see what I am doing wrong

9. Jul 24, 2010

### GreenPrint

It does see

-(x^2 + 16x + 4)/(x^2-x-12)^2

it's right infront...

10. Jul 24, 2010

### GreenPrint

oh crap your right hold up

11. Jul 24, 2010

### GreenPrint

it's suppose to be 12 let me see I still don't see how your suppose to get 3 critical points

12. Jul 24, 2010

### GreenPrint

I still get this as a derivitive

-(x^2 + 16x + 4)/(x^2 - x - 12)^2

13. Jul 24, 2010

### GreenPrint

critical values are 2(-4 +/- sqrt(15) )

still need a third one

It's the same :O

14. Jul 24, 2010

### vela

Staff Emeritus
That's correct. In your first post, however, you wrote
which is slightly different. It's missing negative sign on the quadratic term.

15. Jul 24, 2010

### GreenPrint

Ok I'm sorry please forgive me but I still don't know how to find the third value and don't see how to setting the denominator to zero only yields the vertical asymptotes I need the third critical value in the foruth quardrant of the graph how do I find this

16. Jul 24, 2010

### vela

Staff Emeritus
Why do you think there's a third critical point?

17. Jul 24, 2010

### GreenPrint

Inspect the graph there is one on the left side after it crosses and yes passes through y=1 it drops down and then goes back up in the fourht quadrant no?

18. Jul 24, 2010

### vela

Staff Emeritus
Yes, there's one there, and there's one in the middle section between the two vertical asymptotes. Those are the only two as far as I can see, and they correspond to the two roots you found.

19. Jul 24, 2010

### GreenPrint

ok but put in 2(-4 + sqrt(15)) you get the point in the middile
put in 2(-4 - sqrt(15)) you get the point on the left sorry

20. Jul 24, 2010

### GreenPrint

so then how do I get the point on the right?