# Thread: Is this function consistent with the Extreme Value Theorem?

1. ## Is this function consistent with the Extreme Value Theorem?

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 continuous on 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?

2. ## Re: Is this function consistent with the Extreme Value Theorem?

That is correct. The Extreme Value Theorem says that if f (a real-valued function of a real variable) is continuous on a closed bounded interval, then it achieves its maximum and minimum value. Since your interval is not closed, the theorem says nothing about f, so any behavior is consistent with the theorem.

Your function achieves both maximum and minimum. If the interval were (-1,1), it would achieve its minimum but not its maximum. And if the interval were (0,1) it would achieve neither.

- Hollywood