Finding Absolute Extrema for a Rational Function with Imaginary Solutions

[QuarkCharmer]

Homework Statement

Find the absolute max/min of the function on it's interval.
$$f(x)=\frac{x^2-4}{x^2+4}$$
Interval: [-4,4]

Homework Equations

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

 

[Ivan92]

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

 

[QuarkCharmer]

Ah yeah. Thanks! So the other critical point is 0 then.

 

[Dick]

Your algebra is faulty. The x terms don't cancel. And you probably don't care about imaginary solution in the denominator.

 

[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?

 

[Dick]

$$2x^3+8x-2x^3+8x=0$$
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?

Correct. The imaginary solutions are irrelevant. They aren't in the real interval [-4,4].

 

[QuarkCharmer]

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.