
#1
Jun1811, 11:24 PM

P: 1,035

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^24}{x^2+4}[/tex] Interval: [4,4] 2. Relevant equations 3. The attempt at a solution [tex]f(x)=\frac{x^24}{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^24)}{(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+8x2x^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 TI89 just says "not a function or program, error"? edit: fixed 



#2
Jun1811, 11:28 PM

P: 181

Hint Hint: 2 negatives make a positive. Look at your expansion again :)




#4
Jun1811, 11:31 PM

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P: 25,167

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.




#5
Jun1811, 11:37 PM

P: 1,035

[tex]2x^3+8x2x^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? 



#6
Jun1811, 11:39 PM

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