# Maximums and Minimums

by QuarkCharmer
Tags: maximums, minimums
 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. $$f(x)=\frac{x^2-4}{x^2+4}$$ Interval: [-4,4] 2. Relevant equations 3. The attempt at a solution $$f(x)=\frac{x^2-4}{x^2+4}$$ 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: $$f'(x) = \frac{2x(x^2+4)-2x(x^2-4)}{(x^2+4)^2}$$ Setting the numerator to zero should find where the derivative is equal to zero, but that expands out to this: $$2x^3+8x-2x^3+8x$$ 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
 P: 181 Hint Hint: 2 negatives make a positive. Look at your expansion again :)
 P: 1,035 Ah yeah. Thanks! So the other critical point is 0 then.
HW Helper
Thanks
P: 24,462

## 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.
 P: 1,035 $$2x^3+8x-2x^3+8x=0$$ from the numerator, works out to be: $$16x=0$$ 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?
HW Helper
 Quote by QuarkCharmer $$2x^3+8x-2x^3+8x=0$$ from the numerator, works out to be: $$16x=0$$ 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?