Find the absolute maximum and absolute minimum values of f on the given interval.
f(x)= (x^2-4)/(x^2+4), [-4,4]
Maxima and minima are found where f'(x) = 0
f'(x) = 2x/(x^2 + 4) - 2x(x^2 - 4)/(x^2 + 4)^2
f'(x) = 16x/(x^2 + 4)^2
A turning point is found when x = 0, giving f'(x) = 0
This turning point is at the point (0, -1)
As x -> ꝏ, x^2 – 4 -> x^2
As x -> ꝏ, x^2 + 4 -> x^2
Therefore, as x -> ꝏ, (x^2 – 4)/(x^2 + 4) -> x^2/x^2 = 1
As x -> -ꝏ, (x^2 – 4)/(x^2 + 4) -> x^2/x^2 = 1
i.e. As x -> ±ꝏ, (x^2 – 4)/(x^2 + 4) -> 1
So, the line x = 1 is a horizontal asymptote for the function f(x) = (x^2 – 4)/(x^2 + 4)
Although it is not a local maximum, f(x) has a maximum value of 1 as x -> ±ꝏ
The local minimum would be f(x) = -1 at x = 0