What are the maximum and minimum values of the function on the interval ? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why?
I think I have the right answer, but I want to make sure it is reasonably sound.
According to the Extreme Value Theorem:
If is continuous on a closed interval [a,b], then attains both an absolute maximum value M and an absolute minimum value m in [a,b]. That is, there are numbers and in [a,b] with , and for every other x in [a,b].
So, it seems to me that the EVT asserts the existence of both an absolute max and min value if it is (1) continuous and (2) on a closed interval. However, it doesn't discount the existence of these values if any one of the conditions was not upheld. My argument is that the function on [-1,1) contains both an absolute maximum and minimum value. To be consistent with EVT, due to the function's even symmetry, we can choose to redefine our bounds to [-1,0]—thereby, making it a closed interval, and the outcome would be consistent with the original bounds.
In mathematical terms,
; for f(x) is an even function satisfying .
f(-1) is greater than or equal to all other values of x on the interval [-1,0] and consequently [-1,1), therefore it is an absolute maximum value.
f(0) is the absolute minimum value.
Does this sound reasonable?