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Maximums and Minimums

by QuarkCharmer
Tags: maximums, minimums
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QuarkCharmer
#1
Jun18-11, 11:24 PM
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1. The problem statement, all variables and given/known data
Find the absolute max/min of the function on it's interval.
[tex]f(x)=\frac{x^2-4}{x^2+4}[/tex]
Interval: [-4,4]

2. Relevant equations

3. The attempt at a solution
[tex]f(x)=\frac{x^2-4}{x^2+4}[/tex]

I basically want to find all the critical points, so I set the denominator to zero and found a critical point to be where x = 2i, and x = -2i.

Then I took the derivative of the function as so:

[tex]f'(x) = \frac{2x(x^2+4)-2x(x^2-4)}{(x^2+4)^2}[/tex]

Setting the numerator to zero should find where the derivative is equal to zero, but that expands out to this:
[tex]2x^3+8x-2x^3+8x[/tex]
Which is basically zero anyway. I was going to try to factor out something from the numerator and denominator (if possible) to cancel it out so I could get something, but I thought I would lose solutions doing that. So here is where I am confused? I tried graphing it, thinking that it would just return the line y=0, but my TI-89 just says "not a function or program, error"?

edit: fixed
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Ivan92
#2
Jun18-11, 11:28 PM
P: 181
Hint Hint: 2 negatives make a positive. Look at your expansion again :)
QuarkCharmer
#3
Jun18-11, 11:29 PM
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Ah yeah. Thanks! So the other critical point is 0 then.

Dick
#4
Jun18-11, 11:31 PM
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Maximums and Minimums

Your algebra is faulty. The x terms don't cancel. And you probably don't care about imaginary solution in the denominator.
QuarkCharmer
#5
Jun18-11, 11:37 PM
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[tex]2x^3+8x-2x^3+8x=0[/tex]
from the numerator, works out to be:
[tex]16x=0[/tex]
x=0

I was not sure what to do with the imaginary solutions, essentially the only critical point found is at x=0 then I would imagine. Leaving me to check the endpoints and x=0 so find the absolutes correct?
Dick
#6
Jun18-11, 11:39 PM
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Quote Quote by QuarkCharmer View Post
[tex]2x^3+8x-2x^3+8x=0[/tex]
from the numerator, works out to be:
[tex]16x=0[/tex]
x=0

I was not sure what to do with the imaginary solutions, essentially the only critical point found is at x=0 then I would imagine. Leaving me to check the endpoints and x=0 so find the absolutes correct?
Correct. The imaginary solutions are irrelevant. They aren't in the real interval [-4,4].
QuarkCharmer
#7
Jun18-11, 11:51 PM
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Alright, I think I got it now. This was the only one in the set of problems that had the possibility of an imaginary solution, and I really wasn't sure if that counted or not.

Thanks for the help.


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