PROBLEM:

What are the maximum and minimum values of the function $\displaystyle y=|x|$ on the interval $\displaystyle -1{\leq}x<1$? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why?

ATTEMPT:

I think I have the right answer, but I want to make sure it is reasonably sound.

According to the Extreme Value Theorem:

If $\displaystyle f$ is continuouson a closed interval[a,b], then $\displaystyle f$ attains both an absolute maximum value M and an absolute minimum value m in [a,b]. That is, there are numbers $\displaystyle x_1$ and $\displaystyle x_2$ in [a,b] with $\displaystyle f(x_1)=m, f(x_2)=M$, and $\displaystyle m{\leq}f(x){\leq}M$ 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 $\displaystyle f(x)=|x|$ 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,

$\displaystyle \lim_{x\to\1^-}|x|=f(-1)=1$; for f(x) is an even function satisfying $\displaystyle f(x)=f(-x)$.

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?