absolute maximum of (x^2-4)/(x^2+4)
the critical numbers i got was x=0,-2.59, and 3.09
the absolute minimum is f(0)=-1
what is the absolute maximum:
is it both f(-4)=(f(4))=.6?
or can there only be one
The absolute maximum must occur where the deriviative is zero. But you also have to use othe tests to show that it not a minimum or only a relative maximum.
The derivative is [(x^2+4)(2x) - (x^2-4)(2x)]/(x^2+4)^2
=16 x/(x^2+4)^2
That only equals zero when x = 0. That happens to be the absolute minimum. There is no absolute maximum. The function approaches +1 as x goes to plus or minus infinity.
To find the absolute maximum of the function f(x) = (x^2 - 4) / (x^2 + 4), you first need to find the critical numbers where the derivative is either equal to zero or does not exist. This is because the maximum or minimum of a function can occur at these points.
You have correctly found the critical numbers as x = 0, -2.59, and 3.09. Next, you need to evaluate the function at these critical numbers and at the endpoints of the interval you are considering.
You mentioned that f(0) = -1. This gives you the minimum value of the function, not the maximum. To find the maximum, you need to evaluate the function at the remaining critical numbers and the endpoints.
In this case, you have x = -4 and x = 4 as the endpoints. To find the values of f(-4) and f(4), substitute these values into the function:
f(-4) = (-4^2 - 4) / (-4^2 + 4) = 0.6
f(4) = (4^2 - 4) / (4^2 + 4) = 0.6
Both f(-4) and f(4) have the same value of 0.6, which means that they are both candidates for the absolute maximum. In this case, there can be multiple points that give the same maximum value.
Therefore, the absolute maximum of the function is f(-4) = 0.6 (at x = -4) and f(4) = 0.6 (at x = 4).